id
stringlengths 5
9
| image
imagewidth (px) 876
1.02k
| question
stringlengths 48
392
| answer
stringlengths 93
1.05k
| type
stringclasses 2
values |
|---|---|---|---|---|
1_2_1
|
As shown in the figure, Circle M and Circle N both pass through points A and B, and point N is on Circle M. Point C is a point on major arc AE (point C does not coincide with A or B). The extension of AC intersects Circle N at point P. Connect AB, BC, BP. If ∠APB=30°, AB=3, then the length of MN is
|
Auxiliary lines: Connect MN, AN, BN. Draw MD⊥AN from point M to D, as shown in the figure:
∵ Both circle M and circle N pass through points A and B, ∠APB = 30°, AB = 3,
∴ ∠ANB = 2∠P = 60°,
Also ∵ AN = BN,
∴ △ABN is an equilateral triangle,
∴ AN = AB = 3,
∴ △ABN is inscribed in circle M,
∴ Point M is the circumcenter of the equilateral △ABN,
∴ MN bisects ∠ANB, MD perpendicularly bisects AN,
∴ ∠MND = 30°, DN = 1/2 AN = 1.5,
∵ MD ⊥ AN,
∴ ∠MDN = 90°,
∴ cos ∠MND = DN / MN,
∴ MN = DN / cos ∠MND = 1.5 / cos 30° = √3.
|
calculation
|
|
2_1_1
|
As shown in the figure, in △ABC, ∠ACB=90°, ∠ABC=67.5°, D is the midpoint of AB, and DE⊥AB intersects AC at point E. If BC=2, then the length of AC is
|
Auxiliary line: Connect BE.
Because ∠ACB=90° and ∠ABC=67.5°,
Therefore ∠A=180°-90°-67.5°=22.5°,
Because D is the midpoint of AB, and DE⊥AB intersects AC at point E,
Therefore DE is the perpendicular bisector of AB,
Therefore EB=EA,
Therefore ∠A=∠ABE=22.5°,
Therefore ∠BEC=∠A+∠ABE=45°,
Therefore ∠CBE=180°-90°-45°=45°=∠BEC,
Therefore BC=CE=2,
Therefore BE=√2BC=2√2=EA,
Therefore AC=CE+EA=2+2√2,
Thus, choose: C.
|
calculation
|
|
6_7_2
|
In isosceles triangle ABC, AB=AC, point D is a point on side BC, draw circle O with AD as diameter, intersecting AB and AC at points E and F respectively. When point D is the midpoint of arc EF, (2) Given that the radius of circle O is 6, sin∠ABD=3/5, find the length of BE.
|
∵ Point D is the midpoint of arc EF,
∴ arc DE = arc DF,
∴ ∠BAD = ∠CAD,
∵ AB = AC,
∴ AD ⊥ BC,
∵ AD is the diameter of circle O,
∴ ∠AED = 90°,
∴ ∠BAD + ∠ABD = ∠BAD + ∠ADE = 90°,
∴ ∠ABD = ∠ADE,
∴ sin ∠ABD = sin ∠ADE = 3/5,
∵ The radius of circle O is 6,
∴ AD = 12,
In Rt△ABD, sin ∠ABD = 3/5,
∴ AD/AB = 12/AB = 3/5, solving for AB: AB = 20,
In Rt△ADE, sin ∠ADE = 3/5,
∴ AE/AD = AE/12 = 3/5, solving for AE: AE = 7.2,
∴ BE = AB - AE = 12.8.
|
calculation
|
|
7_6_2
|
As shown in the figure, given that in △ABC, D is the midpoint of AC, draw DE⊥AC through point D, intersecting BC at point E, draw AF//BC through point A, intersecting DE at point F, connect AE and CF. (2) If CF=2, ∠FAC=30°, ∠B=45°, find the length of AB.
|
Because AF // BC,
Therefore ∠FAD = ∠ECD.
Because D is the midpoint of AC, DE ⊥ AC,
Therefore ∠FDA = ∠EDC, AD = CD.
Therefore △ADF ≌ △CDE.
Therefore AF = CE.
Therefore quadrilateral AECF is a parallelogram.
Because DE ⊥ AC,
Therefore parallelogram AECF is a rhombus.
Because AECF is a rhombus,
Therefore AF = CF = 2.
Therefore AD = AF·cos30° = √3.
Therefore AC = 2AD = 2√3.
Draw AM⊥BC through point A.
Therefore AM = AC·sin30° = √3.
Therefore AB = AM/sin45° = √6.
|
calculation
|
|
8_2_1
|
As shown in the figure, in square ABCD, E is the midpoint of AB, FE⊥AB, AF=2AE, FC intersects BD at O, then the measure of ∠COD is
|
Auxiliary lines: Connect DF, BF.
∵ FE⊥AB, AE=EB,
∴ FA=FB,
∵ AF=2AE,
∴ AF=AB=FB,
∴ △AFB is an equilateral triangle,
∵ AF=AD=AB,
∴ Point A is the center of the circumcircle of △DBF,
∴ ∠FDB=1/2∠FAB=30°,
∵ Quadrilateral ABCD is a square,
∴ AD=BC, ∠DAB=∠ABC=90°, ∠ADB=∠DBC=45°,
∴ ∠FAD=∠FBC,
∴ △FAD≌△FBC,
∴ ∠ADF=∠FCB=1/2(180°-60°-90°)=15°,
∴ ∠DOC=∠OBC+∠OCB=60°.
|
calculation
|
|
12_4_1
|
As shown in the figure, in quadrilateral ABCD, AD//BC, ∠C=45°. Construct an isosceles right triangle BAE with AB as a leg, such that vertex E happens to lie on side CD. If CE=6, then the length of AD is
|
Auxiliary line: Draw EF⊥CD through point E, intersecting BC at F.
∵∠C=45°, EF⊥CD, CE=6,
∴EF=CE=6, ∠FED=90°, ∠EFC=45°,
∴∠EBF+∠BEF=∠EFC=45°,
∴∠EFB=135°
∵∆ABE is an isosceles right triangle,
∴∠AEB=45°, BE/AE=√2/1,
∴∠BEF+∠AED=∠FED-∠AEB=90°-45°=45°,
∴∠EBF+∠BEF=∠BEF+∠AED, which means: ∠EBF=∠AED,
∵AD//BC,
∴∠D=180°-∠C=135°
∴∠ADE=∠BFE=135°,
∴∆BEF∽∆EAD,
∴BE/AE=EF/AD, which means: √2/1=6/AD, solving for AD: AD=3√2,
|
calculation
|
|
12_6_2
|
As shown in the figure, AB is tangent to ☉O at point A, radius OC // AB, BC intersects ☉O at point D, connect AD. (2) If the radius of ⊙O is 6, tanB=1/3, find the length of AD.
|
Because AB is tangent to ⊙O at point A,
Therefore ∠OAB=90°.
Because OC∥AB,
Therefore ∠AOC=90°.
Therefore ∠ADC=45°.
Because OC=OA,
Therefore ∠OCA=45°.
Therefore ∠OCA=∠ADC.
Because OC // AB,
Therefore ∠B = ∠OCE.
Because tan B = 1/3,
Therefore tan∠OCE = 1/3.
Because the radius of ⊙O is 6,
Therefore OC = OA = 6.
Therefore tan∠OCE = OE / OC = OE / 6 = 1/3.
Therefore OE = 2.
Therefore AE = OA - OE = 6 - 2 = 4.
In Rt△ABE, tanB = AE / AB = 1/3.
Therefore AB = 12.
Therefore BE = √(4² + 12²) = 4√10. Draw AF ⊥ BC from point A to point F.
Therefore AF = (AE · AB) / BE = (4 × 12) / (4√10) = (6√10) / 5.
From ∠OCA = ∠ADC = 45°,
Therefore △ADF is an isosceles right triangle, AD = √2 · AF = (12√5) / 5.
|
calculation
|
|
14_3_1
|
As shown in the figure, respectively taking sides AB and CD of parallelogram ABCD as legs of right triangles, construct isosceles right triangles ABE and CDF inwards of parallelogram ABCD. Points N and M are taken respectively on the hypotenuses AE of △ABE and CF of △CDF. Connect EM, FN. Quadrilateral EMFN is a square. If the area of parallelogram ABCD is 4, then the area of △ABE is
|
Auxiliary lines: Connect EF, and extend it to intersect AD and BC at points G and H respectively. Connect MN, which intersects EF at point O.
∵ Isosceles Rt△ABE and isosceles Rt△CDF are constructed inside parallelogram ABCD, with sides AB and CD as the right-angle sides respectively. Points N and M are taken on the hypotenuses AE of △ABE and CF of △CDF respectively. EM and FN are connected, and quadrilateral EMFN is a square.
∴ The entire figure is centrally symmetric with point O as the center of symmetry.
∴ S_{quadrilateral ABHG} = S_{quadrilateral DCHG} = \frac{1}{2} S_{quadrilateral ABCD} = \frac{1}{2} \times 4 = 2.
Then the area of △ABE = \frac{1}{2} S_{quadrilateral DCHG} = \frac{1}{2} \times 2 = 1.
Thus, the answer is: 1.
|
calculation
|
|
15_3_1
|
Rhombus ABCD and rectangle EFGD are placed as shown in the figure. Side EF passes through point A, and point G is on side BC. If AB=6, ∠B=60°, DG=4√3, then DE=
|
Auxiliary line: Draw DH⊥BC, intersecting the extension of BC at point H, as shown in the figure,
∵ Quadrilateral ABCD is a rhombus, ∠B=60°, AB=6,
∴ AB∥DC, DC=6,
∴ ∠DCH=∠B=60°,
∴ DH=DC·sin∠DCH=6×√3/2=3√3,
∴ sin∠DGH=DH/DG=3√3/4√3=3/4,
∵ Quadrilateral DEFG is a rectangle,
∴ ∠E=90°, EF∥DG,
∴ ∠EAD=∠ADG,
∵ AD∥BD,
∴ ∠ADG=∠DGH,
∴ ∠EAD=∠DGH,
∴ sin∠EAD=sin∠DGH=3/4,
∴ DE/AD=3/4,
i.e., DE/6=3/4,
Solving for DE, we get DE=9/2,
Therefore, the answer is: 9/2.
|
calculation
|
|
15_5_1
|
As shown in the figure, points A, B, C, D all lie on ☉O, and CB passes through the center. Draw a tangent to ☉O through point A, intersecting the extension of CB at point E. Connect AB, AC, AD, BD. (1) Prove that: ∠ADB=∠EAB;
|
Auxiliary line: Connect OA, as shown in the figure,
∵ AE is a tangent to ⊙O,
∴ OA⊥AE,
∴ ∠OAE=90°,
∴ ∠OAB+∠EAB=90°,
∵ BC is a diameter of ⊙O,
∴ ∠BAC=90°,
i.e., ∠OBA+∠ACB=90°,
∵ OA=OB,
∴ ∠OAB=∠OBA,
∴ ∠EAB=∠ACB,
∵ ∠ACB=∠ADB,
∴ ∠EAB=∠ADB;
|
proving
|
|
17_4_1
|
As shown in the figure, AB is the diameter of ☉O, AC intersects ☉O at point C, the bisector of ∠BAC intersects ☉O at point D, DE⊥AC, with E as the foot of the perpendicular. (1) Prove that: DE is tangent to ☉O.
|
Auxiliary line: Connect OD,
∵ AD is the angle bisector of ∠BAC,
∴ ∠EAD = ∠DAB
∵ OA = OD,
∴ ∠DAB = ∠ODA
∴ ∠EAD = ∠ODA,
∴ AE // OD
∵ DE ⊥ AC,
∴ ∠DEA = 90°,
∴ ∠ODE = 90°
Also ∵ OD is the radius (or D is the endpoint of the radius),
∴ DE is a tangent to ⊙O
|
proving
|
|
20_5_1
|
As shown in the figure, AB is the diameter of ☉O, AC is a chord, point D is the midpoint of arc AB, chord CD intersects AB at point E. Draw a tangent to ☉O through point C, intersecting the extension of AB at point F. (1) Prove that CF=EF;
|
Auxiliary lines: Connect OC, OD.
∵ OC = OD,
∴ ∠OCD = ∠ODC.
∵ CF is a tangent to ⊙O,
∴ ∠OCF = 90°, i.e., ∠FCE + ∠OCD = 90°.
∵ AB is the diameter of ⊙O, and D is the midpoint of AB,
∴ ∠DOE = 90°.
∴ ∠OED + ∠ODC = 90°.
∴ ∠OED = ∠FCE.
Also ∵ ∠OED = ∠FEC,
∴ ∠FCE = ∠FEC.
∴ FC = FE.
|
proving
|
|
21_4_2
|
As shown in the figure, ☉O is the circumcircle of △ABC, AB is the diameter of ☉O, point D is on arc BC, arc AC = arc BD, E is on the extension of BA, ∠CEA=∠CAD. (2) As shown in Figure 2, if ∠CEA=2∠DAB, OA=8, find the length of arc BD.
|
Connect CO, DO,
From (1), we get ∠3=2∠2=2∠4,
∵∠CEA=2∠DAB,
∴∠CEA=∠3,
∵∠ECO=90°,
∴∠3=∠CEA=90°/2=45°,
∴∠4=22.5°,
∵BD=BD,
∴∠DOB=2∠4=45°,
∴The length of BD is: 45×π×8/180=2π.
|
calculation
|
|
25_3_2
|
As shown in the figure, quadrilateral ABCD is inscribed in circle O, AD is the diameter of circle O, the extensions of AD and BC intersect at point E, extend CB to intersect PA at point P, ∠BAP+∠DCE=90°. (2) Connect AC, sinBAC=1/3, BC=2, the length of AD is
|
Auxiliary line: Extend DC to intersect the extension of AB at point F,
∵ AD is the diameter of circle O,
∴ ∠ACD = 90°,
∴ ∠ACF = 180° - ∠ACD = 90°,
∴ △ACF is a right-angled triangle,
∴ sin∠BAC = CF/AF,
∵ Quadrilateral ABCD is inscribed in circle O,
∴ ∠FCB = ∠FAD,
Also ∵ ∠F = ∠F,
∴ △FCB ∼ △FAD,
∴ CB/AD = CF/AF,
∵ sin∠BAC = 1/3, BC = 2,
∴ 2/AD = CF/AF = 1/3,
∴ AD = 6.
|
calculation
|
|
30_2_2
|
As shown in the figure, draw ☉O with side AC of △ABC as diameter, intersecting side BC at point D. Draw CE//AB through point C, intersecting ☉O at point E. Connect AD, DE. ∠B =∠ADE. (2) If tanB=2, CD=3, find the lengths of AB and DE.
|
Solution: ∵ CE // AB,
∴∠BAC = ∠ACE,
∴∠BAC = ∠ACE = ∠ADE,
Q? B ? ADE,
∴∠B = ∠BAC,
∴AC = BC;
Let BD=x,
∵AC is the diameter of ⊙O,
∴∠ADC=∠ADB=90°,
∵tanB=2,
∴AD/BD=2, i.e., AD=2x,
AC=BC=BD+DC=x+3,
According to the Pythagorean theorem, AD²+DC²=AC², i.e., (2x)²+3²=(x+3)²,
Solving for x, we get x₁=2, x₂=0 (discarded),
∴BD=2, AD=4,
According to the Pythagorean theorem, AB=√AD²+BD²=2√5;
|
calculation
|
|
31_2_2
|
As shown in the figure, in right triangle ABC, ∠ACB=90°, draw ☉O with BC as diameter, intersecting side AB at point D, take a point E on arc CD, such that arc BE = arc CD, connect DE, draw ray CE intersecting side AB at point F. (2) If AC=8, cos∠ACF=4/5, find the lengths of BF and DE.
|
Because ∠B=∠BCF, ∠A=∠ACF
Therefore AF=CF, BF=CF,
Therefore AF=BF=1/2 AB,
Because cos∠ACF=cosA=AC/AB=4/5, AC=8,
Therefore AB=10,
Therefore BF=5,
Because BC=√(AB²-AC²)=6,
Therefore sinA=BC/AB=3/5,
Connect CD. Because BC is the diameter of ⊙O,
Therefore ∠BDC=90°,
Therefore ∠B+∠BCD=90°,
Therefore ∠A=∠BCD,
Therefore sin∠BCD=BD/BC=3/5,
Therefore BD=18/5,
Therefore DF=BF-BD=7/5,
Because ∠FDE=∠BCE, ∠B=∠BCE,
Therefore ∠FDE=∠B,
Therefore DE//BC,
Therefore △FDE∽△FBC,
Therefore DE/BC=DF/BF,
Therefore DE=42/25.
|
calculation
|
|
32_4_1
|
As shown in the figure, AB is the diameter of ☉O, C is a point on the circle, D is the midpoint of minor arc BC, a tangent to ☉O passing through point D intersects the extension of AC at point P, and intersects the extension of AB at point F, AD intersects BC at point E. (1) Prove that BC//PF;
|
Auxiliary line: Connect OD. ∵ D is the midpoint of minor arc BC, ∵ D is the midpoint of minor arc BC, ∴ CD = BD, ∴ OD ⊥ BC. Also, ∵ F is a tangent to ⊙O, ∴ OD ⊥ PF, ∴ BC ∥ PF;
|
proving
|
|
38_4_1
|
As shown in the figure, quadrilateral ABCD is inscribed in ☉O, BD is the diameter of ☉O. Draw AE ⊥ CD through point A, intersecting the extension of CD at point E. DA bisects ∠BDE. (1) Prove that AE is tangent to ☉O;
|
∵ OA = OD,
∴ ∠ODA = ∠OAD.
∵ DA bisects ∠BDE,
∴ ∠ODA = ∠EDA.
∴ ∠OAD = ∠EDA,
∴ EC // OA.
∵ AE ⊥ CD,
∴ OA ⊥ AE.
∵ Point A is on ⊙O,
∴ AE is tangent to ⊙O.
|
proving
|
|
40_3_2
|
In right triangle ABC, ∠BAC=90°, D is the midpoint of BC, E is the midpoint of AD. Draw AF//BC through point A, intersecting the extension of CE at point F. (2) If AB=2, ∠AFB=60°, find the length of CF.
|
Auxiliary line: Draw FG⊥CB intersecting the extension of CB at point G, then ∠G=90°.
∵ Quadrilateral ADBF is a rhombus,
∴ AF=BF=AD=BD, ∠ADB=∠AFB=60°,
∴ △ADB and △AFB are both equilateral triangles,
∴ CD=BD=BF=AB=2, ∠ABF=∠ABD=60°,
∴ ∠GBF=180°-∠ABF-∠ABD=60°,
∴ ∠BFG=30°,
∴ BG=\frac{1}{2}BF=1, FG=\sqrt{BF^2+BG^2}=\sqrt{3},
∴ CG=CD+BD+BG=2+2+1=5,
∴ CF=\sqrt{FG^2+CG^2}=\sqrt{(\sqrt{3})^2+5^2}=2\sqrt{7},
∴ The length of CF is 2\sqrt{7}.
|
calculation
|
|
42_4_2
|
As shown in the figure, in △ABC, points D and E are the midpoints of AB and AC respectively, extend DE to point F such that EF=DE, connect CF. (2) If △ABC is an equilateral triangle, BC=6, find the area of △CEF.
|
∵BC=6,
∴EF=DE=1/2BC=3,
∵△ABC is an equilateral triangle,
∴BG=1/2AB=3, ∠B=∠C=∠CAB=60°,
∴AG=√(6²-3²)=3√3,
∵DE//BC,
∴∠ADE=∠B=60°, ∠AED=∠C=60°,
∴△ADE is an equilateral triangle,
∴DH=1/2DE=3/2,
∴AH=3√3/2,
∴HG=AG-AH=3√3/2,
∴The area of △CEF=1/2EF·HG=1/2×3×3√3/2=9√3/4.
|
calculation
|
|
45_4_2
|
As shown in the figure, △ABC is inscribed in ☉O, AE and BD are altitudes to sides BC and AC of △ABC, respectively, chord GH of ☉O passes through points D and E, connect CG, CH, BG, AH, given that ∠BAD and ∠BED are supplementary. (2) Prove that CG=CH;
|
∵ ∠BAD and ∠BED are supplementary,
∴ points A, B, E, D are concyclic,
∴ ∠EAD=∠CBD, ∠AEH=∠ABD,
∴ ∠AGH+∠GAE=∠ABH+∠DBH,
∵ ∠AH=∠AH,
∴ ∠ABH=∠AGH,
∴ ∠GAE=∠DBH,
∴ ∠GAE+∠EAD=∠DBH+∠CBD,
∴ ∠GAC=∠CBH,
∵ ∠CH=∠CH,
∴ ∠CGH=∠CBH,
∵ ∠CG=∠CG,
∴ ∠GAC=∠GHC,
∴ ∠GHC=∠HGC, i.e., △CGH is an isosceles triangle,
∴ CG=CH;
|
proving
|
|
47_5_1
|
As shown in the figure, AB is the diameter of circle O, C, D are two points on circle O, EC is a tangent to circle O, and EC⊥AE, with foot E. Connect AC intersecting BD at point F. (1) Prove that AC bisects ∠EAB;
|
Auxiliary line: Connect OC.
∵ AE⊥CE,
∴ ∠AEC = 90°,
∴ ∠EAC + ∠ACE = 90°,
∵ EC is the tangent to ⊙O,
∴ ∠OCE = 90°,
∴ ∠ACO + ∠ACE = 90°,
∴ ∠ACO = ∠EAC,
∵ OA = OC,
∴ ∠ACO = ∠CAO,
∴ ∠EAC = ∠CAO,
∴ AC bisects ∠EAB;
|
proving
|
|
52_4_1
|
As shown in the figure, given that AB is the diameter of ☉O, point H is on ☉O, E is the midpoint of arc HB. Draw EC⊥AH through point E, intersecting the extension of AH at point C. Connect AE. Draw EF⊥AB through point E at point F. (1) Prove that CE is tangent to ☉O;
|
Auxiliary line: Connect OE.
∵ AO = OE,
∴ ∠EAO = ∠AEO.
∵ HE = EB,
∴ ∠HAE = ∠EAO.
∴ ∠AEO = ∠HAE.
∵ ∠HAE + ∠CEA = 90°,
∴ ∠AEO + ∠CEA = 90°.
∴ ∠CEO = 90°.
And point E is on the circle,
∴ CE is a tangent to ⊙O.
|
proving
|
|
53_3_2
|
In right triangle ABC, ∠BAC=90°, D is the midpoint of BC, draw AE//BC through point A, and AE=BD, connect BE. (2) Connect CE. If AB=2, ∠AEB=60°, find the length of CE.
|
Auxiliary line: Draw EF⊥CB from point E, intersecting the extension of CB at point F.
∵ Quadrilateral ADBE is a rhombus,
∴ AE = BE,
∵ ∠AEB = 60°,
∴ △AEB is an equilateral triangle,
∵ AB = 2,
∴ BE = AB = 2,
∴ BD = DC = BE = 2,
∵ AE // BC,
∴ ∠EBF = ∠AEB = 60°,
In Rt△BEF, ∠F = 90°, ∠EBF = 60°, BE = 2,
∴ ∠BEF = 90° - 60° = 30°,
∴ BF = 1/2 BE = 1,
∴ EF = √(BE² - BF²) = √(2² - 1²) = √3,
∴ CF = BD + DC + FB = 5,
∴ In Rt△CEF, CE = √(CF² + EF²) = √(5² + (√3)²) = 2√7.
|
calculation
|
|
53_4_2
|
As shown in the figure, AB is the diameter of ☉O, points C, E are on ☉O, ∠CAB=2∠EAB, point F is on the extension of segment AB, and ∠AFE=∠ABC. (2) If BF=1, sin∠AFE=4/5, find the length of BC.
|
Solution: Let the radius of ⊙O be x, then OF=x+1,
∵∠AFE=∠ABC, sin∠AFE=4/5,
∴sin∠ABC=4/5,
In Rt△OEF, ∠OEF=90°, sin∠AFE=4/5,
∴OE/OF=4/5, which means x/(x+1)=4/5,
Solving for x gives x=4,
Upon verification, x=4 is the solution to the equation,
∴The radius of ⊙O is 4, then AB=8,
In Rt△ABC, ∠C=90°, sin∠ABC=4/5, AB=8,
∴AC=AB·sin∠ABC=32/5,
∴BC=√(AB²-AC²)=24/5.
|
calculation
|
|
54_1_1
|
As shown in the figure, PA and PB are tangents to circle O at points A and B respectively, C is a point on the minor arc AB, if ∠ACB=130°, then ∠P=
|
Construction: Take any point D on the major arc AB, connect AD, DB, OA, OB.
∵ Quadrilateral ADBC is a cyclic quadrilateral,
∴ ∠D + ∠ACB = 180°.
∴ ∠D = 180° - 130° = 50°.
∵ ∠AOB = 2∠D,
∴ ∠AOB = 100°.
∵ PA and PB are tangents to ⊙O at points A and B,
∴ OA ⊥ AP, OB ⊥ PB.
∴ ∠P + ∠AOB = 180°.
∴ ∠P = 180° - 100° = 80°.
Thus, the answer is: 80°.
|
calculation
|
|
54_4_2
|
As shown in the figure, triangle ABC is inscribed in circle O. The bisector of the exterior angle BAD of triangle ABC intersects circle O at point P (point A is on the arc between P and C). Connect PB, PC. (2) If BC=8, COS∠BAC=3/5, find the length of PB.
|
Auxiliary line: Draw CE⊥PB through point C at E, then ∠PEC=∠BEC=90°,
∵ Quadrilateral ACBP is an inscribed quadrilateral of ⊙O,
∴ ∠PBC = ∠PAD,
∵ AP is the bisector of ∠BAD,
∴ ∠PAD = ∠PAB,
∴ ∠PBC = ∠PAB,
∵ ∠PAB = ∠PCB,
∴ ∠PBC = ∠PCB,
∴ PB = PC;
∵ ∠BAC=∠BPC, cos∠BAC=3/5,
∴ cos∠BPC=3/5,
In Rt△PEC, cos∠BPC=PE/PC=3/5,
∴ Let PE=3x, PC=5x,
According to the Pythagorean theorem, CE=4x,
From PB=PC=5x,
∴ BE=PB-PE=2x,
In Rt△BEC, BC=8,
According to the Pythagorean theorem, BE²+CE²=BC²,
∴ (2x)²+(4x)²=8²,
∴ x=4√5/5 or x=-4√5/5 (discarded),
∴ PB=5x=4√5;
|
calculation
|
|
68_3_1
|
As shown in the figure, AB is the diameter of ☉O, line CD is tangent to ☉O at point C. Connect AC. If ∠ACD=50°, then the measure of ∠BAC is
|
Auxiliary line: Connect OC,
Because line CD is tangent to ⊙O,
Therefore OC⊥CD,
Therefore ∠OCD = 90°,
Because ∠ACD = 50°,
Therefore ∠OCA = 40°,
Because OA = OC,
Therefore ∠BAC = ∠OCA = 40°,
|
calculation
|
|
71_3_1
|
As shown in the figure, in square ABCD, point E is the midpoint of AB, point F is on DE. Connect BF, CF. If BC=BF, ∠DCF=a, then ∠BFE must be equal to
|
Auxiliary line: Extend DE and CB to intersect at point G, as shown in the figure:
Because ABCD is a square,
Therefore ∠EAD = ∠EBG = 90°,
Because E is the midpoint of AB,
Therefore AE = EB,
In △AED and △BEG,
∠EAD = ∠EBG = 90°,
AE = EB,
∠AED = ∠BEG,
Therefore △AED ≅ △BEG (ASA),
Therefore AD = BG = BC = BF,
Therefore ∠G = ∠BFG, ∠BCF = ∠BFC,
Because ∠G + ∠BFG + ∠BCF + ∠BFC = 180°,
Therefore ∠GFC = ∠GFB + ∠BFC = 90°,
Because ∠BCD = 90°, ∠DCF = α,
Therefore ∠BCF = ∠BCD - ∠DCF = 90° - α,
Because BC = BF,
Therefore ∠BFC = ∠BCF = 90° - α,
Therefore ∠BFE = ∠EFC - ∠BFC = α,
|
calculation
|
|
71_6_2
|
In △ABC, ∠ACB=90°, AC=BC. Point D is a moving point on side BC. Draw CE⊥AD from point C, intersecting AB at point E. (2) As shown in the figure, point F is a point on BD. Connect EF and extend it to intersect the extension of AD at point G. If point P is the midpoint of AD, CP=DG, and ∠G=2∠CAD, prove that CE+EF=2FG;
|
Auxiliary line: Draw QB⊥CB, intersecting the extension of CE at point Q.
∵ ∠ACB=90°, and P is the midpoint of AD,
∴ PC=PA=DP=1/2AD,
∴ ∠CAD=∠PCA,
∴ ∠DPC=∠CAD+∠PCA=2∠CAD,
∵ CP=DG, ∠G=2∠CAD,
∴ DG=DP, ∠G=∠DPC,
∵ ∠CDP=∠FDG,
∴ △CDP≌△FDG (ASA),
∴ PC=FG=AP=DP=DG, ∠GFD=∠GDF=∠PDC=∠PCD,
∴ AD=2FG,
Draw QB⊥CB, intersecting the extension of CE at point Q,
∴ ∠Q+∠QCB=90°,
∵ CE⊥AD,
∴ ∠PDC+∠QCB=90°,
∴ ∠Q=∠PDC,
∵ ∠BFE=∠GFD,
∴ ∠Q=∠BFE,
∵ ∠ABC=45°,
∴ ∠QBE=45°=∠EBF,
∵ BE=BE,
∴ △BQE≌△BFE (AAS),
∴ EF=EQ,
∵ AC=CB, ∠Q=∠ADC, ∠ACD=∠CBQ=90°,
∴ △ACD≌△CBQ (AAS),
∴ AD=CQ,
∴ CQ=CE+EQ=CE+EF=AD=2FG,
Thus, CE+EF=2FG;
|
proving
|
|
78_3_1
|
As shown in the figure, quadrilateral ABCD is an inscribed quadrilateral of ☉O, point O is inside quadrilateral ABCD. Draw a tangent to ☉O through point C intersecting the extension of AB at point P. Connect OA, OB. If ∠AOB=140°, ∠BCP=35°, then the measure of ∠ADC is
|
Auxiliary line: Connect OC,
∵ OA = OB = OC, ∠AOB = 140°,
∴ ∠OAB = ∠OBA = 1/2 (180° - ∠AOB) = 20°, ∠OCB = ∠OBC,
∵ CP is a tangent,
∴ ∠OCP = 90°, i.e., ∠OCB + ∠BCP = 90°,
∵ ∠BCP = 35°,
∴ ∠OBC = ∠OCB = 55°,
∴ ∠ABC = ∠ABO + ∠OBC = 75°,
∵ Quadrilateral ABCD is an inscribed quadrilateral of ⊙O,
∴ ∠ADC = 180° - ∠ABC = 105°,
|
calculation
|
|
82_2_1
|
As shown in the figure, in △ABC, ∠ACB=90°. Point O is a point on side AC. Draw a circle with O as the center and OC as the radius, tangent to AB at point D. Connect CD. (1) Prove that: ∠ABC=2∠ACD;
|
Auxiliary line: Connect OD,
∵ AB is a tangent,
∴ OD ⊥ AB,
∴ ∠ODA = 90°,
∴ ∠A + ∠AOD = 90°,
∵ ∠ACB = 90°,
∴ ∠ABC + ∠A = 90°,
∴ ∠AOD = ∠ABC,
∵ ∠AOD = 2∠ACD,
∴ ∠ABC = 2∠ACD.
|
proving
|
|
83_2_2
|
As shown in the figure, AB is the diameter of ☉O, AC is a chord, D is a point on arc AC, P is a point on the extension of AB. Connect AD, DC, CP. (2) If ∠ACP=∠ADC, the radius of ☉O is 3, CP=4, find the length of AP.
|
Auxiliary line: Connect OC,
∵ BC = BC,
∴ ∠BDC = ∠BAC,
∵ AB is the diameter of O,
∴ ∠ADB = 90°,
∴ ∠ADC = ∠ADB + ∠BDC,
∵ ∠BAC = ∠BDC,
∴ ∠ADC = 90° + ∠BAC,
∴ ∠ADC - ∠BAC = 90°,
∴∠ACP-∠BAC=90°,
∵OA=OC,
∴∠BAC=∠ACO,
∴∠ACP-∠ACO=90°,
∴∠OCP=90°,
∵ The radius of ⊙O is 3,
∴AO=OC=3,
In Rt△OCP, OP²=OC²+CP²,
∵CP=4,
∴OP²=3²+4²=25,
∴OP=5,
∴AP=AO+OP=8,
|
calculation
|
|
86_1_1
|
As shown in the figure, in △ABC, AB=AC. ⊙O with AB as diameter intersects BC at point D, intersects the extension of line segment CA at point E, and connect BE. (1) Prove that: BD=CD;
|
Auxiliary line: Connect AD, as shown in the figure:
∵ AB is the diameter of ⊙O,
∴ AD ⊥ BC,
Also ∵ AB = AC,
∴ △ABC is an isosceles triangle,
∴ AD is the perpendicular bisector of BC,
∴ BD = CD.
|
proving
|
|
107_4_1
|
As shown in the figure, AB is the diameter of ☉O, BC intersects ☉O at point D, point E is the midpoint of arc BD, connect AE, intersecting BC at point F, ∠ACB=2∠EAB. (1) Prove that AC is tangent to ☉O;
|
Auxiliary line: Connect AD, as shown in the figure.
∵ E is the midpoint of BD,
∴ DE = BE,
∴ ∠EAB = ∠EAD,
∵ ∠ACB = 2∠EAB,
∴ ∠ACB = ∠DAB,
∵ AB is the diameter of ⊙O,
∴ ∠ADB = 90°,
∴ ∠DAC + ∠DAB = 90°,
∴ ∠DAC + ∠ACB = 90°, i.e., ∠BAC = 90°,
∴ AC ⊥ AB,
∵ OA is the radius of ⊙O,
∴ AC is the tangent to ⊙O;
|
proving
|
|
107_4_2
|
As shown in the figure, AB is the diameter of ☉O, BC intersects ☉O at point D, point E is the midpoint of arc BD, connect AE, intersecting BC at point F, ∠ACB=2∠EAB. (2) If BF=6, cosC=3/5, find the length of AB.
|
Auxiliary line: Connect AD. Draw FH⊥AB from point F to H. And AC⊥AB.
Because QE is the midpoint of BD,
Therefore DE = BE.
Therefore ∠EAB = ∠EAD.
Because ∠ACB = 2∠EAB,
Therefore ∠ACB = ∠DAB.
Because AB is the diameter of ⊙O,
Therefore ∠ADB = 90°.
Therefore ∠DAC + ∠DAB = 90°.
Therefore ∠DAC + ∠ACB = 90°, i.e., ∠BAC = 90°.
Therefore FH∥AC.
Therefore ∠C = ∠BFH.
Therefore cosC = cos∠BFH = FH/BF = 3/5.
Given FB = 6,
Therefore FH = 18/5.
Given ∠EAB = ∠EAD, ∠ADF = 90°, ∠AHF = 90°,
Therefore DF = FH = 18/5.
Therefore DB = 48/5.
Given ∠C + ∠ABC = 90° = ∠ABC + ∠BAD,
Therefore ∠BAD = ∠C.
Therefore cosC = cos∠BAD = AD/AB = 3/5.
Let AD = 3x, AB = 5x.
Therefore BD = √(AB² - AD²) = 4x = 48/5.
Therefore x = 12/5.
Therefore AB = 5x = 12.
|
calculation
|
|
109_3_2
|
As shown in the figure, AB is the diameter of ☉O, point C is on ☉O. If chord CD bisects ∠ACB, intersecting AB at point E, draw a tangent line CF to ☉O through point C, intersecting the extension of AB at point F. (2) Connect BD. If ∠CDB=30° and BF=2, find the length of the radius of ☉O.
|
(2) Solution: Connect BD,
∵ AB is the diameter,
∴ ∠ACB = 90°,
∵ BC = BC,
∴ ∠CDB = ∠CAB = 30°,
∴ ∠ABC = 60°,
∵ OB = OC,
∴ △OBC is an equilateral triangle,
∴ ∠BOC = ∠OCB = ∠ABC = 60°,
∵ In Rt△COF, cos∠COF = OC/OF = 1/2,
Let OC = OB = r, then OF = r + 2,
∴ r/(r + 2) = 1/2,
∴ r = 2,
Upon verification, r = 2 is the solution to the fractional equation and satisfies the problem conditions,
That is, the radius of ⊙O is 2.
|
calculation
|
|
110_4_2
|
As shown in the figure, AB is the diameter of ☉O. Draw a tangent AM to ☉O through point A. C is a point on the semicircle AB (not coinciding with points A, B). Connect AC. Draw CD⊥AB through point C at point E. Connect BD and extend it to intersect AM at point F. (2) If the radius of ☉O is 5, and AC=8, find the length of DF.
|
Solution: Connect AD
∵ AM is a tangent to ⊙O
∴ ∠BAM = 90°
∵ CD ⊥ AB at point E
∴ ∠CEA = 90°
∴ CD∥AF
∴ ∠CDB = ∠AFB
∵ ∠CDB = ∠CAB
∴ ∠CAB = ∠AFB
∵ CD⊥AB at point E, AB is the diameter of ⊙O
∴ CE = DE
∴ AB is the perpendicular bisector of CD
∴ AC = AD = 8
∵ The radius of ⊙O is 5
∴ AB = 10
∴ BD = 6
∵ AB is the diameter of ⊙O
∴ ∠BDA = 90°
∵ ∠BAD = ∠AFB
∴ tan∠BAD = tan∠AFB
∴ AD/BD = DF/AD
∴ AD² = DF·BD
∴ DF = 32/3
|
calculation
|
|
112_5_2
|
As shown in the figure, AB is the diameter of ☉O, AC is a chord, line EF passes through point C, AD⊥EF at point D, ∠DAC=∠BAC. (2) Prove that: AC²=AD·AB;
|
∵ AB is the diameter of ⊙O, AD⊥EF,
∴ ∠BCA = ∠ADC = 90°.
∵ ∠DAC = ∠BAC,
∴ △ACB ∽ △ADC.
∴ AD / AC = AC / AB.
∴ AC² = AD * AB.
|
proving
|
|
113_4_2
|
As shown in the figure, AB is the diameter of ☉O, point C is on ☉O. Connect AC, BC. D is the midpoint of AC. Draw a tangent to ☉O through point C, intersecting ray OD at point E. (2) If EC is extended to intersect AB at point F, if DE=4, sinF=3/5, find the area of ☉O.
|
∵ In Rt△OCF, sin F = OC / OF = 3 / 5,
∴ Let OC = OB = 3k, OF = 5k,
∴ BF = OF - OB = 2k,
∵ D is the midpoint of AC, AO=CO,
∴ OD⊥AC, ∠AOD=∠COD,
∵ AC=AC,
∴ ∠CBA=1/2∠AOC,
∴ ∠CBA=∠COD=∠AOD,
∴ BC ∥ OE,
∴ △FCB ∽ △FEO,
∴ BC / OE = BF / OF = 2 / 5,
∴ Let BC = 2m, OE = 5m,
∵ D is the midpoint of AC, O is the midpoint of AB,
∴ OD is the midline of △ABC,
∴ OD = 1/2 BC = m, DE = OE - OD = 4m,
∵ DE = 4,
∴ m = 1, OD = 1, OE = 5,
∵ OD ⊥ AC, OC ⊥ CE,
∴ ∠OCE = ∠ODC = 90°,
∴ △ODC ∽ △OCE,
∴ OC² = OD × OE = 5,
∴ The area of ⊙O, S = π × OC² = 5π.
|
calculation
|
|
114_3_2
|
As shown in the figure, in parallelogram ABCD, AC, BD intersect at point O, and AO=BO. (2) The angle bisector DE of ∠ADB intersects AB at point E. When AD=3, tan∠CAB=3/4, find the length of AE.
|
Auxiliary line: Draw EG⊥BD through point E at point G, as shown in the figure:
∵ Quadrilateral ABCD is a parallelogram,
∴ AC=2AO, BD=2BO,
∵ AO=BO,
∴ AC=BD,
∴ Parallelogram ABCD is a rectangle.
∴ EA⊥AD,
∵ DE is the angle bisector of ∠ADB,
∴ EG=EA,
∵ AO=BO,
∴ ∠CAB=∠ABD,
∵ AD=3, tan∠CAB=3/4,
∴ tan∠CAB=tan∠ABD=3/4=AD/AB,
∴ AB=4,
∴ BD=√(AD²+AB²)=√(3²+4²)=5, sin∠CAB=sin∠ABD=AD/BD=3/5,
Let AE=EG=x, then BE=4-x,
In Rt△BEG, ∠BGE=90°,
∴ sin∠ABD=x/(4-x)=3/5,
Solving for x: x=3/2,
|
calculation
|
|
115_4_2
|
As shown in the figure, ☉O is the circumcircle of △ABC, AB is the diameter of ☉O. Through a point D on BC, draw DE⊥AB at point E. Through point C, draw a tangent to ☉O, intersecting the extension of ED at point F. (2) If D is the midpoint of BC, the radius of ☉O is 5, and cos∠CDF=3/5, find the length of CF.
|
∵ CF is tangent to ⊙O,
∴ ∠OCF=∠OCB+∠DCF=90°,
∴ ∠DCF=90°-∠OCB,
∵ EF⊥AB at E,
∴ ∠FEB=90°,
In Rt△EBD,
∴ ∠EBD+∠EDB=90°,
∴ ∠EDB=90°-∠EBD,
Also ∵ BC and EF intersect at point D,
∴ ∠CDF=∠EDB,
∴ ∠CDF=90°-∠EBD,
∵ OC=OB,
∴ ∠EBD=∠OCB,
∴ ∠DCF=∠CDF
∵ AB is the diameter of ⊙O,
∴ ∠ACB=∠ACO+∠OCB=90°,
Also ∵ ∠OCF=∠DCF+∠OCB=90°,
∴ ∠ACO=∠DCF,
∵ OA=OC, ∠DCF=∠CDF,
∴ ∠ACO=∠OAC=∠DCF=∠CDF,
∴ △FCD∽△OCA,
∵ cos∠CDF=3/5, ∠OAC=∠DCF=∠CDF,
∴ cos∠OAC=3/5,
∵ The radius is 5,
∴ AB=10,
In Rt△ABC, ∠ACB=90°,
∴ cos∠OAC=AC/AB=3/5,
∴ AC=6, BC=8,
∵ D is the midpoint of BC,
∴ CD=BD=4,
∵ △FCD∽△OCA,
∴ CD/CA=CF/CO,
That is 4/6=CF/5,
∴ CF=10/3.
|
calculation
|
|
121_2_1
|
As shown in the figure, PA and PB are tangent to ☉O at points A and B, respectively. C is a moving point on the major arc AB. If ∠P=50°, then ∠ACB=
|
Auxiliary line: Connect OA, OB, as shown in the figure.
∵ PA and PB are tangent to ⊙O respectively
∴ ∠OAP = ∠OBP = 90°
∴ ∠AOB = 360° - ∠OAP - ∠OBP - ∠P = 130°
∵ Arc AB = Arc AB
∴ ∠ACB = 1/2 ∠AOB = 65°
|
calculation
|
|
127_4_2
|
As shown in the figure, in △ABC, AB=AC. ☉O with AB as diameter intersects BC at point D, DE⊥AC, with foot E. (2) If ∠C=30°, CD=2√3, find the length of BD.
|
Auxiliary line: Connect AD,
∵ AB is the diameter of ⊙O,
∴ AD⊥BC,
∵ AB=AC, CD=2√3
∴ BD=CD=2√3.
|
calculation
|
|
131_5_1
|
As shown in the figure, AB is the diameter of ☉O, point C is a point on the circle, CD⊥AD at point D, AD intersects ☉O at point F, connect AC. If AC bisects ∠DAB, draw FG⊥AB from point F at point G, intersecting AC at point H. (1) Prove that CD is tangent to ☉O;
|
Auxiliary line: Connect OC.
∵ OA = OC,
∴ ∠CAO = ∠ACO,
∵ AC bisects ∠DAB,
∴ ∠DAC = ∠OAC,
∴ ∠DAC = ∠ACO,
∴ AD ∥ OC,
∵ CD ⊥ AD,
∴ OC ⊥ CD,
∵ OC is the radius of ⊙O,
∴ CD is a tangent to ⊙O;
|
proving
|
|
138_4_1
|
As shown in the figure, given that BC is the diameter of ☉O, point D is the midpoint of arc CE. Draw DG//CE through point D, intersecting the extension of BC at point A. Connect BD, intersecting CE at point F. (1) Prove that AD is tangent to ☉O;
|
Auxiliary lines: Connect OD, BE.
∵ Point D is the midpoint of CE,
∴ CD = ED,
∴ ∠CBD = ∠EBD,
∵ OB = OD,
∴ ∠ODB = ∠CBD,
∴ ∠ODB = ∠EBD,
∴ OD // BE,
∵ BC is the diameter of ⊙O,
∴ ∠CEB = 90°,
∴ CE ⊥ BE,
∴ OD ⊥ CE,
∵ AD // CE,
∴ AD ⊥ OD,
∵ OD is the radius of ⊙O,
∴ AD is a tangent to ⊙O.
|
proving
|
|
141_5_2
|
As shown in the figure, in △ABC, ∠ABC=∠ACB. ☉O with AC as diameter intersects AB and BC at points M and N respectively. Point P is on the extension of AB, and 2∠BCP=∠BAC. (2) If BC=3√2 and cosBCP=√30/6, find the distance from point B to AC.
|
Auxiliary line: Connect AN. Draw BQ⊥AC from point B, intersecting AC at point Q, as shown in the figure.
Because AC is the diameter of ⊙O,
Therefore ∠ANC=∠ANB=90°,
i.e., AN⊥BC.
Because ∠ABC=∠ACB,
Therefore AB=AC.
By the property of isosceles triangles (three lines coincide),
AN bisects ∠BAC, and AN bisects BC.
Therefore ∠BAN=∠CAN=1/2∠BAC, BN=CN=1/2BC=3√2/2.
Because ∠BCP=1/2∠BAC,
Therefore ∠BAN=∠CAN=∠BCP.
Therefore cos∠BAN=cos∠CAN=cos∠BCP=√30/6.
Because in Rt△ACN, cos∠CAN=AN/AC=√30/6,
Therefore AN=AC×cos∠CAN.
Let AC=x, AN=√30/6x.
In Rt△ACN, according to the Pythagorean theorem, AN²+CN²=AC².
Therefore 9/2+5/6x²=x²,
i.e., 1/6x²=9/2,
Therefore x²=27, x=3√3.
i.e., AC=3√3, AN=3√10/2.
According to the triangle area formula, S△ABC=1/2×AC×BQ=1/2×BC×AN.
Therefore 1/2×3√3×BQ=1/2×3√2×3√10/2.
Solving for BQ gives BQ=√15.
i.e., the distance from point B to AC is √15.
|
calculation
|
|
144_4_1
|
As shown in the figure, in YABCD, AB=5, AD=3√2, ∠A=45°. Find the length of diagonal BD;
|
Auxiliary line: Connect BD. Draw DF⊥AB through D at F, as shown in the figure:
∵ In ▱ABCD, AD=3√2, ∠A=45°,
∴ AF=DF=AD/√2=3,
∵ AB=5,
∴ BF=AB-AF=5-3=2,
In Rt△BDF, ∠BFD=90°, DF=3, BF=2, then BD=√(DF²+BF²)=√(3²+2²)=√13;
|
calculation
|
|
147_3_1
|
As shown in the figure, AB is the diameter of ☉O, △ACD is inscribed in ☉O, ∠ACD=∠CAB, AD=2, AC=4, then the radius of ☉O is
|
Auxiliary line: Connect BC.
∵ ∠ACD = ∠CAB,
∴ AD = BC,
∴ AD = BC = 2,
∵ AB is the diameter of ⊙O,
∴ ∠ACB = 90°,
∵ AC = 4,
∴ AB = √(AC² + BC²) = √(4² + 2²) = 2√5,
∴ The radius of ⊙O is 1/2 AB = √5.
|
calculation
|
|
147_5_1
|
As shown in the figure, point E is a point on side BC of square ABCD, ☉O is the circumcircle of △ABE, intersecting AD at point F, G is a point on CD, and ∠DGF=∠AEB. (1) Prove that FG is tangent to ☉O;
|
Auxiliary line: Connect OF,
∵ Quadrilateral ABCD is a square,
∴ ∠D=∠ABE=90°,
∴ ∠DGF+∠DFG=∠AEB+∠BAE,
∵ ∠DGF=∠AEB,
∴ ∠DFG=∠BAE,
∵ OA=OF,
∴ ∠OAF=∠DFA,
∵ ∠BAE+∠OAF=90°,
∴ ∠DFG+∠OFA=90°,
∴ ∠OFG=90°, i.e., OF⊥FG,
∴ FG is a tangent to ⊙O;
|
proving
|
|
149_4_1
|
As shown in the figure, in △ABC, ∠C=90°. Point D is a point on side AB. Circle O with BD as diameter is tangent to side AC at point E, and intersects side BC at point F. Draw EH⊥AB from point E at point H. Connect BE. (1) Prove that: BC=BH;
|
Auxiliary line: Connect OE, as shown in the figure.
∵ AC is a tangent,
∴ OE⊥AC,
∴ ∠AEO=90°,
∵ ∠C=90°,
∴ OE∥BC,
∴ ∠1=∠3,
∵ OB=OE,
∴ ∠2=∠3,
∴ ∠1=∠2,
∵ EH=EC,
In Rt△BEH and Rt△BEC,
$$\begin{cases}BE=BE\\EH=EC\end{cases}$$
∴ Rt△BEH≌Rt△BEC(HL),
∴ BC=BH;
|
proving
|
|
153_4_1
|
As shown in the figure, in right triangle ABC, ∠ACB=90°, CD is the median to the hypotenuse. Draw ☉O with CD as diameter, intersecting BC at point E. Draw EF⊥AB through point E, with F as the foot of the perpendicular. (1) Prove that EF is tangent to ☉O;
|
Auxiliary line: Connect OE. Draw circle O with CD as diameter, intersecting BC at point E.
∴ OA = OE,
∴ ∠OCE = ∠OEC,
∵ In Rt△ABC, ∠ACB = 90°, and CD is the median to the hypotenuse,
∴ CD = AD = BD,
∴ ∠OCE = ∠B,
∴ ∠OEC = ∠B,
∴ OE ∥ AB,
∵ EF ⊥ AB,
∴ OE ⊥ EF,
Also ∵ OE is the radius of ⊙O,
∴ EF is the tangent to ⊙O;
|
proving
|
|
154_4_2
|
As shown in the figure, AB is the diameter of ☉O, C and D are two points on ☉O, OC//BD, and chord AD intersects BC and OC at E and F, respectively. (2) If CE=1 and EB=3, find the radius of ☉O.
|
Auxiliary line: Connect AC, as shown in the figure,
∵ AB is the diameter of the circle,
∴ ∠ADB = 90°,
∵ OC // BD,
∴ ∠AFO = ∠ADB = 90°,
∴ OC ⊥ AD,
∴ AC = CD
∴ ∠CAD = ∠ABC,
∵ ∠ECA = ∠ACB,
∴ △ACE ∽ △BCA,
∴ AC² = CE × CB, i.e., AC² = 1 × (1 + 3),
∴ AC = 2 (negative value discarded),
∵ AB is the diameter of the circle,
∴ ∠ACB = 90°,
∴ AB = √(AC² + BC²) = √(2² + 4²) = 2√5,
∴ The radius of circle O is √5.
|
calculation
|
|
155_3_1
|
As shown in the figure, in isosceles △ABC, AB=AC, ☉O with AB as diameter intersects BC at point D, DE⊥AC, with foot E, the extension of ED intersects the extension of AB at point F. (1) Prove that EF is a tangent to 0;
|
Auxiliary line: Connect OD.
Because AB = AC,
Therefore ∠ABC = ∠ACB,
Because OB = OD,
Therefore ∠ABC = ∠ODB,
Therefore ∠ACB = ∠ODB,
Therefore OD ∥ AC,
Because DE ⊥ AC,
Therefore DE ⊥ OD,
i.e., EF ⊥ OD,
Because OD is the radius of ⊙O,
Therefore EF is a tangent to ⊙O.
|
proving
|
|
160_2_1
|
As shown in the figure, points A, B, C are three points on ☉O, and quadrilateral ABCO is a parallelogram. OF⊥OC intersects circle O at point F. Connect AF, then ∠BAF equals
|
Auxiliary line: Connect OB, as shown in the figure.
Because quadrilateral ABCO is a parallelogram,
Therefore OC=AB, and OA=OB=OC,
Therefore OA=OB=AB,
Therefore △AOB is an equilateral triangle,
Therefore OF⊥OC, OC//AB,
Therefore OF⊥AB,
Therefore ∠BOF=∠AOF=30°,
By the inscribed angle theorem, ∠BAF = 1/2 ∠BOF = 15°,
Thus, choose: B.
|
calculation
|
|
163_1_1
|
As shown in the figure, AB is the diameter of ☉O, C and D are two points on ☉O. If ∠BCD=40°, then the measure of ∠ABD is
|
Auxiliary line: Connect AC, as shown in the figure,
∵ AB is the diameter of ⊙O,
∴ ∠ACB = 90°,
∴ ∠ACD = 90° - ∠BCD = 90° - 40° = 50°,
∴ ∠ABD = ∠ACD = 50°.
|
calculation
|
|
163_2_1
|
As shown in the figure, quadrilateral ABCD is an inscribed quadrilateral of ☉O, ∠ADC=150°, chord AC=2, then the radius of ☉O is equal to
|
Auxiliary line: Connect OA, OC.
Because quadrilateral ABCD is an inscribed quadrilateral of ⊙O,
Therefore ∠ADC + ∠ABC = 180°.
Because ∠ADC = 150°,
Therefore ∠ABC = 30°.
Therefore ∠AOC = 2∠ABC = 60°.
Because OA = OC,
Therefore △OAC is an equilateral triangle.
Therefore OA = AC = 2.
That is, the radius of ⊙O is 2.
|
calculation
|
|
163_3_1
|
As shown in the figure, in △ABC, AB=AC, ☉O with AB as diameter intersects BC at point D, DE∠AC intersects the extension of BA at point E, intersects AC at point F. (1) Prove that: DE is tangent to ☉O;
|
∵ AB = AC,
∴ ∠ABC = ∠ACB,
∵ OB = OD,
∴ ∠OBD = ∠ODB,
∴ ∠ODB = ∠ACB,
∴ AC || OD,
∴ ∠DFC = ∠ODF,
∵ DE ⊥ AC,
∴ ∠DFC = ∠ODF = 90°,
∴ OD ⊥ DE,
∴ DE is the tangent to ⊙O;
|
proving
|
|
164_1_1
|
As shown in the figure, in △ABC, ∠ACB=70°. The incircle ☉O of △ABC is tangent to AB and BC at points D and E respectively. Connect DE. The extension of AO intersects DE at point F. Then the measure of ∠AFD is
|
Auxiliary line: Connect OD, OE, OB. OB intersects ED at point G.
Because ∠ACB=70°,
Therefore ∠CAB+∠CBA=110°,
Because point O is the incenter of △ABC,
Therefore ∠OAB+∠OBA=55°,
Therefore ∠AOB=125°,
Because OE=OD, BD=BE,
Therefore OB is the perpendicular bisector of DE,
Therefore ∠OGE=90°,
Therefore ∠AFD=∠AOB-∠OGF=125°-90°=35°.
|
calculation
|
|
164_5_1
|
Given: ☉O is the circumcircle of △ABC, the bisector of ∠ACB intersects ☉O at point D, connect AD, BD. Prove that: AD=BD;
|
Auxiliary line: Connect OA, OB, OD,
∵ OD bisects ∠ACB,
∴ ∠ACD = ∠BCD,
∵ ∠AOD = 2∠ABD, ∠BOD = 2∠BCD, ∠ABD = ∠ACD,
∴ ∠AOD = ∠BOD,
∴ AD = BD.
|
proving
|
|
166_4_1
|
Given: As shown in the figure, triangle ABC is inscribed in circle O, AM is the diameter of circle O, AM bisects ∠BAC. Prove that: AM⊥BC;
|
Auxiliary lines: Connect BM, CM, as shown in the figure.
Proof: Connect BM, CM, as shown in the figure,
∵ AM is the diameter of ⊙O,
∴ ∠ABM=∠ACM=90°.
∵ AM bisects ∠BAC,
∴ ∠BAM=∠CAM, and AM=AM,
∴ △ABM≌△ACM,
∴ AB=AC, BM=CM,
∴ AM is the perpendicular bisector of BC, i.e., AM⊥BC.
|
proving
|
|
168_4_1_1
|
As shown in the figure, in circle O, diameter AB is perpendicular to chord CD at point G, connect AD, draw CF perpendicular to AD at F through point C, intersecting AB at point H, intersecting circle O at point E, connect DE, prove that ∠E=2∠C;
|
∵ AB is the diameter of ⊙O, AB ⊥ CD,
∴ BC = BD, ∠BAD + ∠ADG = 90°
∴ ∠CAB = ∠BAD = 1/2 ∠CAD = 1/2 ∠CED,
∵ AF ⊥ CE,
∴ ∠ECD + ∠ADG = 90°,
∴ ∠ECD = ∠BAD,
∴ ∠E = 2∠DCE;
|
proving
|
|
168_4_3_1
|
As shown in the figure, in circle O, diameter AB is perpendicular to chord CD at point G. Connect AD. From point C, draw CF perpendicular to AD at F, intersecting AB at point H, and intersecting circle O at point E. Connect DE, connect BE, intersecting AD and CD at points M and N respectively. When OH=2OG and HF=√10, find the length of segment EN.
|
∵ OH = 2OG,
∴ Let OG = x, then OH = 2x,
∴ HG = OH + OG = 3x,
∵ AB⊥CD, CE⊥AD,
∴ ∠ECD + ∠CHG = ∠ECD + ∠CDF = 90°,
∴ ∠CHG = ∠ADC,
Also ∵ ∠ADC = ∠B,
∴ ∠CHG = ∠B,
∴ CH = CB
∵ AB ⊥ CD,
∴ BG = GH = 3x,
∴ OB = BG + OG = 4x,
∴ OC = 4x, AB = 8x, AH = 2x,
∵ ∠CHB = ∠AHE, ∠CBH = ∠CEA, and ∠CHB = ∠CBH,
∴ ∠AHE = ∠CEA,
∴ AE = AH = 2x,
∴ In △AHE, BE = √(AB² - AE²) = 2√15x,
In Rt△OGC, CG = √(OC² - OG²) = √15x,
In Rt△HGC, CH = √(CG² + GH²) = 2√6x,
∵ BE = BC = m,
∴ ∠BAD = ∠DCE,
∴ sin∠BAD = sin∠DCE, i.e., HF/AH = HG/CH,
∴ √10/2x = 3x/2√6x,
∴ x = 2√15/3,
∴ BE = 2√15x = 20, BG = 3x = 2√15, AB = 8x = 16√15/3,
∵ ∠ABE = ∠GBN, ∠BGN = ∠AEB = 90°,
∴ △BGN ∽ △BEA,
∴ BN/AB = BG/BE,
∴ BN = (BG · AB)/BE = (2√15 · 16√15/3)/20 = 8,
∴ EN = BE - BN = 12.
|
calculation
|
|
170_3_1
|
In ABCD, diagonals AC and BD intersect at point O. AD=5, AC=8, BD=6. Extend BC to point E. Connect OE intersecting CD at point F. If 2∠E=∠ACD, then the length of segment OF is
|
Auxiliary lines: Take the midpoint M of CD, connect OM, draw ON⊥BE from point O at point N.
1. In quadrilateral ABCD, diagonals AC and BD intersect at point O, AD=5, AC=8, BD=6.
2. ∵ OA=OC=4, OB=OD=3, BC=AD=5.
3. ∴ OB² + OC² = BC².
4. ∴ OB⊥OC.
5. ∴ Quadrilateral ABCD is a rhombus.
6. ∴ ∠ACB=∠ACD.
7. ∵ E is the midpoint of CD, ∴ ∠E=1/2∠ACD.
8. ∴ ∠E=1/2∠ACB, i.e., ∠ACB=2∠E.
9. ∵ ∠ACB=∠E+∠COE.
10. ∴ ∠E=∠COE.
11. ∵ CE=OC=4.
12. ∵ Point M is the midpoint of CD, and point O is the midpoint of BD.
13. ∴ OM is the midline of △BCD.
14. ∴ OM=1/2BC=5/2, OM∥BC.
15. ∴ △FOM∽△FEC.
16. ∴ OF/EF=OM/CE=5/8.
17. ∴ OF/OE=OF/(OF+EF)=5/13.
18. ∵ S△BOC=1/2OB·OC=1/2BC·ON.
19. ∴ ON=OB·OC/BC=12/5.
20. In Rt△OON, CN=√(OC²-ON²)=√(4²-(12/5)²)=16/5.
21. ∴ EN=CN+CE=16/5+4=36/5.
22. In Rt△ONE, OE=√(ON²+EN²)=√((12/5)²+(36/5)²)=12√10/5.
23. ∴ OF=5/13OE=5/13×12√10/5=12√10/13.
|
calculation
|
|
174_5_1_1
|
Given: quadrilateral ABCD is inscribed in ☉O, BC=CD. Prove that: ∠BAD=2∠CBD;
|
Auxiliary line: Connect AC, as shown in the figure:
∵ BC = CD,
∴ BC = CD,
∴ ∠BAC = ∠CAD = ∠BDC,
∵ ∠BAD = ∠BAC + ∠CAD,
∴ ∠BAD = 2∠CBD.
|
proving
|
|
179_3_2
|
As shown in the figure, O is a point on the diagonal of square ABCD, circle O with O as the center and OC as the radius is tangent to AD at point E, and intersects AC at point F. (2) If the side length of square ABCD is √2+1, find the radius of circle O.
|
Solution: Because AC is the diagonal of square ABCD,
Therefore ∠DAC = 45°.
Because ⊙O is tangent to AD at point E,
Therefore ∠AEO = 90°.
Because ⊙O is tangent to AD at point E, Therefore OE⊥AD,
Therefore ∠AEO=∠AGO=90°.
Because quadrilateral ABCD is a square,
Therefore ∠BAC=∠DAC=45°.
Also, Because AO=AO,
Therefore △AOE≅△AOG(AAS),
Therefore AE = OE. Let AE = OE = OC = OF = R.
In Rt△AEO,
Because AE² + EO² = AO²,
Therefore AO² = R² + R².
Because R > 0, Therefore AO = √2R.
Also, Because the side length of square ABCD is √2 + 1,
In Rt△ADC,
Therefore AC = √(AD² + CD²) = √2(√2 + 1).
Because OA + OC = AC,
Therefore √2R + R = √2(√2 + 1),
Therefore R = √2.
Therefore the radius of ⊙O is √2.
|
calculation
|
|
187_3_1
|
As shown in the figure, in ☉O, CD is a chord of ☉O, diameter AB ⊥ CD, connect AC, OD, ∠A=26°, then the measure of ∠D is
|
Auxiliary line: Connect OC,
∵ CD is a chord of ⊙O, and diameter AB⊥CD,
∴ BC = BD,
∴ ∠COB = ∠BOD,
∵ ∠A = 26°,
∴ ∠COB = 2∠A = 52°,
∴ ∠BOD = 52°,
∴ ∠D = 90° - ∠BOD = 90° - 52° = 38°.
|
calculation
|
|
190_4_3
|
As shown in the figure, in right triangle ABC, ∠BAC=90°. Draw circle A with center A tangent to BC at point D. Circle A intersects AC at point M and intersects AB at point N. Point E is a point on circle A, and arc MN = arc DE. Draw EF perpendicular to BD from point E at point F. EF intersects segment AB at point G. Connect BE. (3) If BE²=BG·BA, find the value of tan∠ABC.
|
Solution: ∵ ☉A is tangent to BC,
∴ ∠ADB = 90°,
∵ EF ⊥ BD,
∴ AD ∥ EF,
∵ MN = DE,
∴ MD = NE,
∴ ∠MAD = ∠NAE,
∵ ∠BAC = 90°,
∴ ∠DAE = 90°,
∵ AD ∥ EF,
∴ ∠AEF = 90°,
∵ AE ⊥ EF,
∴ Line EF is tangent to ☉A.
\( QB \cdot BE^2 = BG \cdot BA \)
\(\therefore BE = \frac{AB}{GB} \)
\(\because \angle ABE = \angle EBG \)
\(\therefore \triangle VABE \sim \triangle VEBG \)
\(\therefore \angle BEF = \angle GAE \)
\(\because \angle CAD = \angle GAE \)
\(\therefore \angle CAD = \angle BEF \)
\(\because \) Quadrilateral \( ADFE \) is a rectangle
\(\therefore AD = EF, \angle ADC = \angle EFB = 90^\circ \)
\(\therefore \triangle VACD \cong \triangle VEFB (ASA) \)
\(\therefore CD = BF \)
Let the radius of circle A be r, CD=BF=x,
∵ Quadrilateral ADFE is a rectangle,
∴ DF=AE=r,
∴ BD=DF+BF=r+x,
∵ ∠ADC=∠CAB=90°,
∴ ∠ABC+∠C=90°, ∠CAD+∠C=90°,
∴ ∠ABC=∠CAD,
∵ ∠ADC=∠ADB=90°,
∴ △ADC∽△BDA,
∴ AD/CD=BD/AD, i.e., r/(r+x)=x/r,
Let x/r=m,
∴ 1+m=1/m,
Solving gives: m=(√5-1)/2 or m=(-1-√5)/2 (discarded),
In Rt△ADC, tan∠CAD=CD/AD=√5-1/2,
∵ ∠CAD=∠ABC,
∴ tan∠ABC=√5-1/2.
|
calculation
|
|
191_4_1
|
As shown in the figure, polygon ABCDEF is a regular hexagon, and the side length AB is 4. Then the area of this regular hexagon is
|
Auxiliary line: Let the diagonals of regular hexagon ABCDEF intersect at point O, and draw OH⊥AB at point H.
Because ∠AOB = 1/6 × 360° = 60°, and OA = OB,
Therefore, △AOB is an equilateral triangle,
Therefore, OA = OB = AB = 4,
Because OH⊥AB,
Therefore, AH = BH = 1/2 × AB = 2,
Therefore, OH = √(OB² - BH²) = √(4² - 2²) = 2√3,
Therefore, S△AOB = 1/2 × AB × OH = 1/2 × 4 × 2√3 = 4√3,
Therefore, the area of the regular hexagon = 6 × S△AOB = 6 × 4√3 = 24√3.
|
calculation
|
|
196_6_2
|
As shown in the figure, in △ABC, ∠A=90°. Points O and D are on BC and AC respectively, and CD·BC=AC·OC. Draw a circle with O as the center and OD as the radius. ☉O passes through point B, intersecting AB and BC at points E and F respectively. (2) If AE=4, CF=5, and the radius of the inscribed circle of △ABC is r, find the value of r.
|
Solution: Draw OH⊥AB through point O, with H as the foot of the perpendicular.
∵CD·BC=AC·OC,
∴CD/CA=CO/CB.
∵∠DCO=∠ACB,
∴△DCO∽△CAB.
∴∠ODC=∠A=90°.
∴OD⊥AC.
Also ∵point D is on ⊙O,
∴AC is tangent to ⊙O at point D.
∴BH=EH.
In quadrilateral AHOD, ∠AHO=∠A=∠ADO=90°,
∴quadrilateral AHOD is a rectangle,
∴OD=HA, OD∥AB,
∴∠B=∠DOC.
Let the radius of ⊙O be R, then OB=OD=OF=R,
∴BH=EH=R-4, OC=R+5.
In △BHO and △DOC,
∵∠B=∠DOC, ∠BHO=∠ODC,
∴△BHO∽△DOC,
∴BH/OD=BO/OC, i.e., (R-4)/R=R/(R+5),
∴R=20, which satisfies the problem conditions;
That is, the radius of ⊙O is 20.
∴BC=45, OD=AH=20, BH=EH=20-4=16,
∴AB=36,
∴AC=√(BC^2-AB^2)=27,
∴S_ABC=1/2×36×27=486,
∵the radius of the inscribed circle of △ABC is r,
∴1/2(AB+BC+AC)r=486,
∴1/2(27+36+45)r=486,
Solving for r: r=9
|
calculation
|
|
200_6_2
|
As shown in the figure, AB and CD are two diameters of ☉O, and AB⊥CD. Point E is a moving point on BD (not coinciding with points B or D). Connect DE and extend it to intersect the extension of AB at point F. Point P is on AF, and ∠PEF=∠DCE. Connect AE and CE, intersecting OD and OB at points M and N respectively. Connect AC. Let the radius of ☉O be r. (2) When ∠DCE=15°, prove that AM=2ME;
|
Because ∠DCE=15°,
Therefore ∠DOE=30°,
Because AB⊥CD, then ∠AOD=90°,
Therefore ∠AOE=120°,
Because OA=OE,
Therefore ∠OAE=∠OEA=30°,
Then AM=2OM,
Also, because ∠DOE=30°=∠OEA,
Therefore OM=ME,
Therefore AM=2ME;
|
proving
|
|
201_6_1
|
As shown in the figure, in right triangle ABC, ∠ABC=90°. O is a point on AC. With O as the center and OA as the radius, draw a circle intersecting AC at F. BD is a tangent to ☉O, and D is the point of tangency. Connect OB intersecting ☉O at M. Given ∠DBO=∠CBO, (1) Prove that BC is a tangent to ☉O.
|
Auxiliary line: Connect OD. Draw OE⊥BC from point O to point E.
Because BD is a tangent to ⊙O,
Therefore OD⊥BD,
Therefore ∠ODB=90°,
Because OE⊥BC,
Therefore ∠OEB=90°,
Therefore ∠ODB=∠OEB,
In △ODB and △OEB,
∠ODB=∠OEB,
∠DBO=∠EBO,
OB=OB,
Therefore △ODB≅△OEB (AAS),
Therefore OD=OE,
That is, OE is the radius of ⊙O,
Therefore BC is a tangent to ⊙O.
|
proving
|
|
205_2_1
|
As shown in the figure. AB is the diameter of ☉O, C and D are two points on ☉O. If ∠BCD=28°, then ∠ABD=
|
Auxiliary line: Connect AD,
∵ AB is the diameter of ⊙O,
∴ ∠ADB = 90°,
∵ BD = BD,
∴ ∠BAD = ∠BCD = 28°,
∵ As shown, AB is the diameter of ⊙O,
∴ ∠ABD = 90° - ∠BAD = 62°.
|
calculation
|
|
206_2_1
|
As shown in the figure, triangle ABC is inscribed in circle O, angle A = 45°, CD is perpendicular to AB at point D. If AB = 8, CD = 6, then the radius of circle O is
|
Because ∠A=45°, CD⊥AB at point D, AB=8, CD=6,
Therefore ∠ACD=∠A=45°, AD=CD=6, BD=AB-AD=8-6=2,
Therefore BC=√(CD²+BD²)=√(6²+2²)=2√10.
Because ∠A=45°,
Therefore ∠COB=90°.
Also, because CO=BO,
Therefore △BCO is an isosceles right triangle.
Therefore CO=BO=BC/√2=(2√10)/√2=2√5.
|
calculation
|
|
210_2_1
|
As shown in the figure, square ABCD has a side length of 4. The sides EB and EC of triangle EBC intersect side AD at points F and G, respectively. If the area of triangle EBG is 6, then the ratio of the lengths of FG to BC is
|
Auxiliary line: Draw EM⊥BC from point E to M, intersecting AD at N.
1. In square ABCD, ∠BAD=∠ABC=90°, AD∥BC,
2. Also ∵EM⊥BC,
3. ∴Quadrilateral ABNM is a rectangle,
4. ∴AB=MN, and the side length of square ABCD is 4,
5. ∴Area of square ABCD = 4×4=16,
6. ∴S_BGC=1/2×16=8,
7. ∵The area of △EBG is 6,
8. ∴S_BCE=8+6=14=1/2×BC×EM,
9. ∴EM=7,
10. ∴EN=3,
11. AD∥BC, EM⊥BC,
12. ∴△EFG∽△EBC, EN⊥AD,
13. ∴FG/BC=EN/EM=3/7, therefore, choose: C.
|
calculation
|
|
212_1_1
|
As shown in the figure, △ABC is inscribed in ☉O, ∠A=45°, BC=2√2, then the length of the minor arc BC is
|
Auxiliary line: Connect OB, OC.
∵∠A=45°,
∴∠BOC=90°,
∵BC=2√2,
∴OB=OC=2,
∴ The length of minor arc BC is (90×π×2)/180=π.
|
calculation
|
|
212_3_2
|
As shown in the figure, AB is the diameter of ☉O, CD is a chord of ☉O, the bisector of ∠BCD intersects ☉O at point E, the extensions of AD and BE intersect at point F, (2) Prove that AB=AF.
|
Auxiliary line: Connect DE.
∵ CE bisects ∠BCD, ∴ ∠BCE = ∠DCE,
∴ ∠BDE = ∠EBD.
∵ AB is the diameter of ⊙O, ∴ ∠BDF = 90°,
∴ ∠DBE + ∠F = 90°,
∠BDE + ∠EDF = 90°, ∴ ∠EDF = ∠F.
∵ ∠EDF + ∠ADE = 180°,
∠ABE + ∠ADE = 180°,
∴ ∠EDF = ∠ABE, ∴ ∠ABF = ∠F,
∴ AB = AF.
|
proving
|
|
213_4_2
|
As shown in the figure, in quadrilateral ABCD, AB=BC, diagonals AC, BD intersect at point O, ∠BAC=∠ADB=60°, point E is a point on BD, BE=AD. Connect CE. (2) If M is the midpoint of side AB, connect DM and extend it to intersect the extension of CB at point N, ∠N=∠ACD, BE=2, MD=3, find the length of MN.
|
Auxiliary line: Draw AG∥NC intersecting the extension of ND at point G.
∵ AB = BC, ∠BAC = 60°,
∴ △ABC is an equilateral triangle.
∴ AC = BC, ∠ACB = 60°.
∵ ∠ADB = 60°,
∴ ∠ACB = ∠ADB = 60°.
In △ADO and △BCO,
∵ ∠AOD = ∠BOC, ∠ADO = ∠BCO = 60°,
∴ ∠DAO = ∠CBO.
∴ ∠CBD = ∠CAD, and BE = AD.
In △ADC and △BEC,
\[
\begin{cases}
BC = AC \\
∠CBE = ∠CAD \\
BE = AD
\end{cases}
\]
∴ △ADC ≅ △BEC (SAS).
∴ CE = CD, ∠BCE = ∠ACD.
∵ ∠BCE + ∠ECO = 60°,
∴ ∠ECO + ∠ACD = 60°.
∴ △DCE is an equilateral triangle.
∵ ∠G=∠N, ∠GAM=∠NBM,
∵ Point M is the midpoint of AB, i.e., MA=MB,
∴ △AMG ≅ △BMN (AAS).
∴ MN=MG.
∵ △DCE is an equilateral triangle,
∴ ∠DEC=∠EBC+∠ECB=60°.
∵ AG∥NC,
∴ ∠GAC=∠GAD+∠CAD=∠ACB=60°, and ∠DAC=∠EBC.
∴ ∠GAD+∠DAC=∠EBC+∠ECB.
∴ ∠GAD=∠ECB=∠DCA=∠N.
∴ ∠AGD=∠N.
∴ ∠GAD=∠G.
∴ DG=DA=BE=2.
∵ MD=3,
∴ MG=MD+DG=3+2=5.
∴ MN=MG=5.
|
calculation
|
|
214_3_1
|
As shown in the figure, point P is the centroid of △ABC, point D is the midpoint of side AC, PE//AC intersects BC at point E, DF//BC intersects EP at point F. If the area of quadrilateral CDFE is 6, then the area of △ABC is
|
Auxiliary line: Connect BD.
Because point P is the centroid of △ABC, and point D is the midpoint of side AC,
Therefore, points B, P, D are collinear, and BP:PD=2:1, S_{BCD}=\frac{1}{2}S_{ABC}.
Because PE//AC,
Therefore, △BEP∽△BCD,
Therefore, BP:PD=2:1,
Therefore, BP:BD=2:3,
Therefore, S_{BEP}:S_{BCD}=4:9,
Therefore, S_{BEP}=\frac{4}{9}S_{BCD},
Therefore, S_{quadrilateral CEPD}=S_{BCD}-S_{BEP}=\frac{5}{9}S_{BCD}.
Because DF//BC,
Therefore, △BEP∽△DFP,
Therefore, BP:PD=2:1,
Therefore, S_{BEP}:S_{DFP}=4,
Therefore, S_{DFP}=\frac{1}{4}S_{BEP}=\frac{1}{4}\times\frac{4}{9}S_{BCD}=\frac{1}{9}S_{BCD},
Therefore, S_{quadrilateral CEPD}=S_{quadrilateral CEPD}+S_{DFP}=\frac{5}{9}S_{BCD}+\frac{1}{9}S_{BCD}=\frac{6}{9}S_{BCD}=6,
Therefore, S_{BCD}=9,
Therefore, S_{ABC}=18.
Thus, choose: C.
|
calculation
|
|
218_4_1
|
As shown in the figure, in quadrilateral ABCD, ∠BCD=90°, diagonals AC, BD intersect at point O. If AB=AC=5, BC=6, ∠ADB=2∠CBD. (1) Find the area of triangle ABC.
|
Auxiliary line: Draw AH⊥BC through point A at point H.
∵∠AHC=∠AHB=90°,
∵AB=AC=5, BC=6,
∴BH=HC=1/2BC=3,
∴AH=√(AC²-CH²)=4,
∴S△ABC=1/2BC·AH=1/2×6×4=12;
Therefore, the answer is: 12;
|
calculation
|
|
219_3_1
|
As shown in the figure, quadrilateral ABCD is inscribed in ☉O, arc AD = arc BD, diagonal AC is the diameter of ☉O. Extend BC to intersect the tangent line through point D at point E. (1) Prove that: DE⊥BE;
|
Auxiliary line: Connect DO and extend it to intersect AB at F,
∵ arc AD = arc BD,
∴ DF ⊥ AB,
∵ DE is tangent to ⊙O,
∴ DF ⊥ DE,
∴ DE ⊥ AB,
∵ AC is the diameter of ⊙O,
∴ BE ⊥ AB,
∴ DE ⊥ BE;
|
proving
|
|
221_2_1
|
As shown in the figure, point E is on side AD of rhombus ABCD, AE=2ED. Connect BE, intersecting the diagonal AC of the rhombus at point F. Take the midpoint G of AB. Connect FG. If AB=5, ∠D=120°, then FG=
|
Auxiliary lines: Connect BD intersecting AC at O. Draw BM∥GF through B intersecting AC at M. Draw MN⊥BC through M, with N on BC.
∵ ABCD is a rhombus, AB=5, ∠D=120°,
∴ AB=BC=AD=CD=5, AC⊥BD, AD∥BC, ∠DAC=∠DCA=30°=∠BCA,
∴ OD=OB=5/2, OA=OC=5√3/2,
∴ AC=5√3,
∵ AD∥BC,
∴ △AEF∽△CBF, and AE=2DE,
∴ EF/BF=AF/CF=AE/BC=2/3,
∴ AF=2√3, CF=3√3,
∵ GF∥BM, G is the midpoint of AB,
∴ AG/BG=AF/MF=1,
∴ AF=FM=2√3, GF=1/2BM,
∴ CM=5√3-2√3-2√3=√3,
And ∠ACB=30°,
∴ MN=√3/2, CN=3/2,
∴ BN=5-3/2=7/2,
∴ BM=√(BN²+MN²)=√13,
∴ GF=√13/2,
Therefore, choose: B
|
calculation
|
|
223_2_2
|
As shown in the figure, point C is a point on the extension of diameter AB of ☉O, CE is tangent to ☉O at point D, AE intersects ☉O at point F, ∠BDC=∠DAE. (2) If EF=2, BD=2√5, find the length of AF.
|
Auxiliary line: Connect DF,
∵ CE is tangent to ⊙O,
∴ ∠ODC = 90°,
∵ AB is the diameter of ⊙O,
∴ ∠ADB = 90°,
∴ ∠ADE + ∠BDC = 90°,
∵ ∠BDC = ∠DAE,
∴ ∠ADE + ∠DAE = 90°,
∴ ∠E = ∠ODC = 90°,
∴ OD ∥ AE,
∴ ∠EAD = ∠ADO,
∵ OA = OD,
∴ ∠ADO = ∠OAD,
∴ ∠OAD = ∠EAD,
∴ arc BD = arc BF;
∴ BD = DF = 2√5,
∴ DE = √(DF² - EF²) = √((2√5)² - 2²) = 4,
∵ ∠ADB = 90°,
∴ ∠OAD + ∠ABD = 90°,
∵ ∠ADE + ∠DAE = 90° and ∠OAD = ∠EAD,
∴ ∠ADE = ∠ABD,
∵ ∠ABD = ∠DFE,
∴ ∠ADE = ∠DFE,
∵ ∠E = ∠E,
∴ △EAD ∽ △EDF,
∴ DE / AE = EF / DE,
∴ AE = DE² / EF = 8,
∴ AF = AE - EF = 6.
|
calculation
|
|
232_2_1
|
As shown in the figure, quadrilateral ABCD is a rhombus, ∠ABC=60°. Extend BC to point E. CM bisects ∠DCE. Draw DF⊥CM through point D, with F as the foot of the perpendicular. If DF=1, then the length of diagonal BD is
|
Auxiliary line: Connect AC, intersecting BD at point O.
∵ Quadrilateral ABCD is a rhombus,
∴ AB = BC, ∠CBO = ∠ABO, OB = OD, AC⊥BD,
∵ ∠ABC = 60°,
∴ ∠OBC = 30°, ∠BCD = 120°,
∴ ∠DCE = 60°,
∵ CM bisects ∠DCE,
∴ ∠DCF = ∠ECF = 30°,
∵ DF = 1, DF⊥CM,
∴ DC = 2DF = 2,
∴ OC = 1/2 CD = 1,
∴ OD = √(CD² - OC²) = √3,
∴ BD = 2OD = 2√3.
|
calculation
|
|
233_1_1
|
As shown in the figure, circle O with radius 1 is tangent to the regular pentagon ABCDE at points A and C. The length of the minor arc AC is
|
Auxiliary lines: Connect OA, OB, OC.
∵ Circle O is tangent to regular pentagon ABCDE at points A and C,
∴ ∠OAE=∠OCD=90°,
∴ ∠AOC=540°-∠E-∠D-∠OAE-∠OCD=144°,
Therefore, the length of the minor arc AC is \(\frac{144π×1}{180}=\frac{4}{5}π\).
|
calculation
|
|
233_3_1
|
As shown in the figure, in △ABC, ∠C=90°, D is a point on side BC, if ∠BAD=45°, CD=5, AC=12, then the length of AB is
|
Auxiliary line: Draw DE⊥AB through D, as shown in the figure.
1. ∵∠BAD=45°, ∠AED=90°,
∴∠BAD=∠ADE=45°, then AE=DE.
2. Let AE=DE=x.
3. In Rt△ADE, AD=√2DE, i.e., 13=√2x.
4. Solving for x gives x=13√2/2, so AE=ED=13√2/2.
5. ∵S△ABD=1/2AB·DE=1/2BD·AC, then AB×13√2/2=12a, solving for AB gives AB=12√2a/13.
6. In Rt△ABC, AB²=BC²+AC², i.e., (12√2a/13)²=(a+5)²+12².
7. That is, 119a²/169-10a-169=0.
8. Solving for a gives a=(10±√(100+4×119×169))/(2×119)=(10±24)/169.
9. Then a=2873/119 (negative value discarded).
10. ∴AB=12√2/13a=12√2/13×2873/119=156√2/7.
|
calculation
|
|
235_1_1
|
As shown in the figure, in △ABC, AB=BC=AC, D is the midpoint of BC, DE⊥AB at point E, then the ratio of the area of △BDE to the area of △ABC is
|
Auxiliary line: Connect AD
∵ AB = BC = AC
∴ △ABC is an equilateral triangle
∴ ∠B = 60°
∵ DE ⊥ AB
∴ ∠BDE = 30°
∴ BE = 1/2 BD
∴ DE = √(BD² - BE²) = √3/2 BD
∴ S△BDE = 1/2 BE × DE = 1/2 × 1/2 BD × √3/2 BD = √3/8 BD²
∵ D is the midpoint of BC
∴ AD ⊥ BC, BD = 1/2 BC = 1/2 AB
∴ AD = √(AB² - AD²) = √3/2 AB
∴ S△ABC = 1/2 BC × AD = 1/2 × 1/2 AB × √3/2 AB = √3/4 AB² = √3/4 (2BD)² = √3 BD²
∴ The ratio of the area of △BDE to the area of △ABC is (√3/8 BD²)/(√3 BD²) = 1/8
|
calculation
|
|
235_5_2
|
Given: As shown in the figure, AB is the diameter of ☉O, AC is a chord, CD is a tangent to ☉O, C is the point of tangency, AD⊥CD at point D. Prove: (2) If the radius of ☉O is 3, AD=2, find tan∠ACD.
|
∵ AB is the diameter of circle O,
∴ ∠ACB = 90°,
∴ ∠ABC + ∠OAC = 90°,
∵ CD is the tangent to circle O,
∴ ∠OCD = 90°,
i.e., ∠OCA + ∠DCA = 90°,
∵ OC = OA,
∴ ∠OAC = ∠OCA,
∴ ∠ABC = ∠ACD,
∵ AD⊥CD,
∴ ∠ADC = ∠ACB = 90°,
Also ∠ABC = ∠ACD,
∴ △ABC ∽ △ACD,
∴ AB/AC = AC/AD,
∴ 6/AC = AC/2,
∴ AC = 2√3,
Solving for AC, we get AC = 2√3,
In Rt△ACD, CD = √(AC² - AD²) = 2√2,
∴ In Rt△ACD, tan∠ACD = AD/CD = √2/2.
|
calculation
|
|
238_5_2_1
|
In △ABC, BA=BC, D is a moving point on side BC. Circle O passing through point A is tangent to side BC at point D, and intersects sides AB and AC at points E and F, respectively. Connect AD. As shown in the figure, AD is the diameter of circle O. Connect EF. If AB=√10, AC=2, find the length of EF.
|
Because AF = AF,
Therefore ∠AEF = ∠ADF.
Because BA = BC,
Therefore ∠BAC = ∠C.
Because BC is tangent to ⊙O at point D,
Therefore AD⊥BC, i.e., ∠ADC = 90°,
Therefore ∠C + ∠DAC = 90°,
Also AD is a diameter,
Therefore ∠AFD = 90°,
Therefore ∠FAD + ∠ADF = 90°,
Therefore ∠C = ∠ADF,
Therefore ∠AEF = ∠BAC, Therefore EF = AF.
Let CD = x,
In Rt△ABD, AD² = AB² - BD²;
In Rt△ACD, AD² = AC² - CD²,
Because AB = BC = √10, AC = 2,
Therefore (√10)² - (√10 - x)² = 2² - x²,
Solving for x, we get x = √10/5. (Note: One can also use the area method to calculate AD = 3√10/5, CD = √10/5)
Because △CDA ∽ △CFD,
Therefore CD/CF = CA/CD, i.e., (√10/5)² = 2CF, solving for CF, we get CF = 1/5.
Therefore AF = 9/5, i.e., EF = AF = 9/5.
|
calculation
|
|
242_2_1
|
As shown in the figure, in square ABCD, O is the midpoint of diagonal AC, E is a point inside the square. Connect BE, BE=BA. Connect CE and extend it, intersecting the bisector of ∠ABE at point F. Connect OF. If AB=2, then the length of OF is
|
Auxiliary line: As shown, connect AF.
∵ Quadrilateral ABCD is a square,
∴ AB = BE = BC, ∠ABC = 90°,
AC = √2AB = 2√2,
∠BEC = ∠BCE,
∴ ∠EBC = 180° - 2∠BEC,
∴ ∠ABE = ∠ABC - ∠EBC = 2∠BEC - 90°,
∵ BF bisects ∠ABE,
∴ ∠ABF = ∠EBF = 1/2∠ABE = ∠BEC - 45°,
∴ ∠BFE = ∠BEC - ∠EBF = 45°,
In △BAF and △BEF,
AB = EB
BF = BF
∴ △BAF ≌ △BEF (SAS),
∴ ∠BFE = ∠BFA = 45°,
∴ ∠AFC = ∠BFA + ∠BFE = 90°,
∵ O is the midpoint of diagonal AC,
∴ OF = 1/2AC = √2.
|
calculation
|
|
242_3_1
|
As shown in the figure, in parallelogram ABCD, AD=2AB, F is the midpoint of AD, draw CE⊥AB, with foot E on segment AB, connect EF, CF, prove that: EF=CF.
|
Auxiliary line: Extend BA and CF to intersect at point G.
∵ Quadrilateral ABCD is a parallelogram,
∴ AB∥CD,
∴ ∠G = ∠FCD,
∵ F is the midpoint of AD,
∴ AF = DF,
In △AFG and △DFC,
∠G = ∠FCD,
∠AFG = ∠DFC,
AF = DF,
∴ △AFG ≌ △DFC (AAS),
∴ GF = CF = 1/2 CG,
∵ CE ⊥ AB, with foot E on segment AB,
∴ ∠CEG = 90°,
∴ EF = 1/2 CG,
∴ EF = CF.
|
proving
|
|
245_5_1
|
As shown in the figure, AD is the diameter of ☉O, AB is a chord of ☉O, OP⊥AD, OP intersects the extension of AB at point P, point C is on OP, satisfying ∠CBP=∠ADB. (1) Prove that BC is tangent to ☉O;
|
Auxiliary line: Connect OB, as shown in the figure.
∵ AD is the diameter of ⊙O,
∴ ∠ABD = 90°,
∴ ∠A + ∠ADB = 90°,
∵ OA = OB,
∴ ∠A = ∠OBA,
∵ ∠CBP = ∠ADB,
∴ ∠OBA + ∠CBP = 90°,
∴ ∠OBC = 180° - 90° = 90°,
∴ BC ⊥ OB,
∴ BC is a tangent to ⊙O;
|
proving
|
|
248_3_2
|
Rhombus ABCD has diagonals AC and BD intersecting at point O. If ∠ABC=45°, draw AM⊥BC at point M from A, intersecting BD at point N. (2) If AB=4, find the length of AN:
|
Auxiliary line: Draw NH⊥AB through N at H,
∵BD bisects ∠ABC,
Also ∵NH⊥AB, AM⊥BC,
∴NM=NH,
Let NM=NH=x,
∵∠ABC=45°,
∴△ABM and △AHN are both isosceles right triangles,
∴NM=NH=x=AH, AH²+HN²=AN², AB²=AM²+BM²,
∴AN=√2x,
∴BM=AM
=x+√2x
=2√2,
Thus x=4-2√2,
∴AN=4√2-4.
|
calculation
|
|
251_1_1
|
As shown in the figure, in circle O with radius 5, AB and CD are two mutually perpendicular chords, their intersection point is P, and AB=CD=6, then the length of OP is
|
Auxiliary lines: Draw OM ⊥ AB at M, ON ⊥ CD at N, connect OB, OD.
1. By the Perpendicular Bisector Theorem, AM = BM = DN = CN = 3
2. By the Pythagorean Theorem, OM = ON = √(BO² - BM²) = √(5² - 3²) = 4
3. ∵ Chords AB and CD are perpendicular to each other,
∴ ∠DPB = 90°
4. ∵ OM ⊥ AB at M, ON ⊥ CD at N,
∴ ∠OMP = ∠ONP = 90°
5. ∴ Quadrilateral MONP is a rectangle,
6. ∵ OM = ON,
∴ Quadrilateral MONP is a square,
7. ∴ OP = 4√2
|
calculation
|
|
251_3_2
|
As shown in the figure, in △ABC, O is a point on AC, draw a circle with O as the center and OC as the radius, which is tangent to BC at point C, draw AD⊥BO from point A, intersecting the extension of BO at point D, and ∠AOD=∠BAD. (2) If BC=8, tan∠ABC=4/3, find the length of AD.
|
∵AD⊥BO at point D,
∴∠D=90°,
∵∠BAD+∠ABD=90°, ∠AOD+∠OAD=90°,
∵∠AOD=∠BAD,
∴∠ABD=∠OAD,
Also, ∵BC is a tangent to ⊙O,
∴AC⊥BC,
∴∠BCO=∠D=90°,
∵∠BCO=∠AOD,
∴∠OBC=∠OAD=∠ABD,
In △BOC and △BOE,
∵∠OBC=∠OBE,
∵∠OCB=∠OEB,
∵BO=BO,
∴△BOC≌△BOE (AAS),
∴OE=OC,
Given that ∠ABC+∠BAC=90°, ∠EOA+∠BAC=90°,
∴∠EOA=∠ABC,
Given that BC=8, tan∠ABC=4/3,
∴AC=BC·tan∠ABC=32/3,
Then AB=√(AC²+BC²)=40/3,
From BE=BC=8,
∴AE=40/3-8=16/3,
Given that tan∠EOA=tan∠ABC=4/3,
∴AE/OE=4/3,
∴OE=4, OB=√(BE²+OE²)=4√5,
Given that ∠ABD=∠OBC, ∠D=∠ACB=90°,
∴∠ABD∽∠OBC,
∴OC/OB=OB/AB, i.e., AD/AB=4√5/40,
∴AD=8√5/3.
|
calculation
|
|
259_1_1
|
Given: As shown in the figure, quadrilateral ABCD is an inscribed square of ☉O. Point P is any point on the minor arc CD different from point C. Then the measure of ∠BPC is equal to
|
Auxiliary line: Connect AC,
∵ Quadrilateral ABCD is an inscribed square of ⊙O,
∴ ∠BAC = 45°,
∴ ∠BPC = ∠BAC = 45°,
Therefore, choose A.
|
calculation
|
|
262_3_1
|
As shown in the figure, given that A, B, C, D, E are five points on ⊙O, the center O is on AD, ∠BCD=110°, then the measure of ∠AEB is
|
Auxiliary line: Connect BD, as shown in the figure.
∵ Quadrilateral ABCD is inscribed in ⊙O,
∴ ∠BCD + ∠BAD = 180°,
∵ ∠BCD = 110°,
∴ ∠BAD = 180° - 110° = 70°,
∵ AD is a diameter,
∴ ∠BAD + ∠ADB = ∠ABD = 90°,
∴ ∠ADB = 90° - ∠BAD = 90° - 70° = 20°,
∵ In the same circle, inscribed angles subtended by the same arc are equal,
∴ ∠AEB = ∠ADB,
∴ ∠AEB = 20°.
Therefore, choose: D.
|
calculation
|
|
262_5_1
|
AB is the diameter of ☉O, C is a point on ☉O. Draw a tangent CE to ☉O through point C. Draw BD ⊥ CE through point B at point D, intersecting ☉O at point F. Connect AF. (1) Prove that ∠ABC=∠DBC;
|
Auxiliary line: Connect OC, intersecting AF at point G.
∵ CE is a tangent to ⊙O,
∴ OC⊥ED,
∵ BD⊥CE,
∴ ∠BDC = ∠OCE = 90°,
∴ OC∥BD,
∴ ∠OCB = ∠DBC,
∵ OC = OB,
∴ ∠OCB = ∠OBC,
∴ ∠ABC = ∠DBC;
|
proving
|
|
263_3_2
|
As shown in the figure, in rhombus ABCD, AB=2, ∠A=60°. Circle D with center D is tangent to AB at point E, and intersects DC at point F. (2) Find the length of the minor arc EF.
|
∵ Quadrilateral ABCD is a rhombus,
∴ AB//CD, AD = AB,
∵ ∠A = 60°,
∴ ∠ADC = 120°, △ADB is an equilateral triangle,
∴ AD = BD, ∠ADB = 60°, ∠CDB = 60°,
∴ ∠EDB = 30°, DE = √3
∴ ∠EDC = 60° + 30° = 90°,
∴ The length of minor arc EF is (90° × π × √3) / 180 = (√3 / 2)π.
|
calculation
|
|
263_4_1
|
As shown in the figure, AB is the diameter of ☉O, point C is on ☉O and does not coincide with points A, B, CD is the tangent to ☉O. From point B, draw BD ⊥ CD at point D, intersecting ☉O at point E. (1) Prove that: point C is the midpoint of arc AE;
|
Auxiliary line: Connect OC and AE, intersecting at point P,
∵ AB is the diameter of ⊙O,
∴ ∠AEB = 90°, i.e., AE⊥BD,
∵ BD⊥CD,
∴ AE∥CD,
∵ CD is a tangent to ⊙O,
∴ OC⊥CD,
∴ OC⊥AE,
∴ AC = CE, i.e., C is the midpoint of AE;
|
proving
|
End of preview. Expand
in Data Studio
README.md exists but content is empty.
- Downloads last month
- 15