11. 转化思想(2025·南京中华中学附中期中)如图,在$\triangle ABC$中,$AB= AC$,$AD是\triangle ABC$的角平分线,$FE是AC$的垂直平分线,交$AD于点F$,连接$BF$. 求证：$AF= BF$.
答案：11.连接CF,
∵AB = AC,AD是△ABC的角平分线，
∴BD = CD,AD⊥BC,即AD是BC的垂直平分线，
∴BF = CF.
∵FE是AC的垂直平分线,
∴AF = CF,
∴AF = BF.
∵AB = AC,AD是△ABC的角平分线，
∴BD = CD,AD⊥BC,即AD是BC的垂直平分线，
∴BF = CF.
∵FE是AC的垂直平分线,
∴AF = CF,
∴AF = BF.
12. (2023·苏州中考)如图,在$\triangle ABC$中,$AB= AC$,$AD为\triangle ABC$的角平分线. 以点$A$为圆心,$AD$长为半径画弧,与$AB$,$AC分别交于点E$,$F$,连接$DE$,$DF$.
(1)求证：$\triangle ADE \cong \triangle ADF$；
(2)若$\angle BAC= 80^\circ$,求$\angle BDE$的度数.
(1)求证：$\triangle ADE \cong \triangle ADF$；
(2)若$\angle BAC= 80^\circ$,求$\angle BDE$的度数.
答案：12.
(1)
∵AD是△ABC的角平分线,
∴∠BAD = ∠CAD.
由作图,知AE = AF.
在△ADE和△ADF中,AE = AF,∠EAD = ∠FAD,AD = AD,
∴△ADE≌△ADF(SAS).
(2)
∵∠BAC = 80°,AD为△ABC的角平分线,
∴∠EAD = $\frac{1}{2}$∠BAC = 40°.
又AE = AD,
∴∠AED = ∠ADE,
∴∠ADE = $\frac{1}{2}$×(180° - 40°) = 70°.
∵AB = AC,AD为△ABC的角平分线，
∴AD⊥BC,
∴∠ADB = 90°,
∴∠BDE = 90° - ∠ADE = 20°.
解后反思 本题考查了全等三角形的判定与性质、角平分线的性质、等腰三角形的性质，熟练掌握全等三角形的判定是解题的关键
(1)
∵AD是△ABC的角平分线,
∴∠BAD = ∠CAD.
由作图,知AE = AF.
在△ADE和△ADF中,AE = AF,∠EAD = ∠FAD,AD = AD,
∴△ADE≌△ADF(SAS).
(2)
∵∠BAC = 80°,AD为△ABC的角平分线,
∴∠EAD = $\frac{1}{2}$∠BAC = 40°.
又AE = AD,
∴∠AED = ∠ADE,
∴∠ADE = $\frac{1}{2}$×(180° - 40°) = 70°.
∵AB = AC,AD为△ABC的角平分线，
∴AD⊥BC,
∴∠ADB = 90°,
∴∠BDE = 90° - ∠ADE = 20°.
解后反思 本题考查了全等三角形的判定与性质、角平分线的性质、等腰三角形的性质，熟练掌握全等三角形的判定是解题的关键
13. 中考新考法 类比探究 如图,在$\triangle ABC$中,$AB= AC$,点$D在BC$上,且$BD= BA$,点$E在BC$的延长线上,且$CE= CA$.
(1)若$\angle BAC= 90^\circ$(如图(1)),求$\angle DAE$的度数；
(2)若$\angle BAC= 120^\circ$(如图(2)),求$\angle DAE$的度数；
(3)当$\angle BAC＞90^\circ$时,探求$\angle DAE与\angle BAC$之间的数量关系,并说明理由.
(1)若$\angle BAC= 90^\circ$(如图(1)),求$\angle DAE$的度数；
(2)若$\angle BAC= 120^\circ$(如图(2)),求$\angle DAE$的度数；
(3)当$\angle BAC＞90^\circ$时,探求$\angle DAE与\angle BAC$之间的数量关系,并说明理由.
答案：13.
(1)
∵AB = AC,∠BAC = 90°,
∴∠B = ∠ACB = 45°.
∵BD = BA,
∴∠BAD = ∠BDA = $\frac{1}{2}$(180° - ∠B) = 67.5°.
∵CE = CA,
∴∠CAE = ∠E = $\frac{1}{2}$∠ACB = 22.5°,
∴∠BAE = 180° - ∠B - ∠E = 112.5°,
∴∠DAE = ∠BAE - ∠BAD = 45°.
(2)
∵AB = AC,∠BAC = 120°,
∴∠B = ∠ACB = 30°.
∵BA = BD,
∴∠BAD = ∠BDA = $\frac{1}{2}$(180° - ∠B) = 75°,
∴∠DAC = ∠BAC - ∠BAD = 45°
∵CA = CE,
∴∠E = ∠CAE = $\frac{1}{2}$∠ACB = 15°,
∴∠DAE = ∠DAC + ∠CAE = 60°.
(3)∠DAE = $\frac{1}{2}$∠BAC.理由如下:
设∠CAE = x、∠BAD = y,
则∠B = 180° - 2y,∠E = ∠CAE = x,
∴∠BAE = 180° - ∠B - ∠E = 2y - x,
∴∠DAE = ∠BAE - ∠BAD = 2y - x - y = y - x,∠BAC = ∠BAE - ∠CAE = 2y - x - x = 2y - 2x,
∴∠DAE = $\frac{1}{2}$∠BAC.
(1)
∵AB = AC,∠BAC = 90°,
∴∠B = ∠ACB = 45°.
∵BD = BA,
∴∠BAD = ∠BDA = $\frac{1}{2}$(180° - ∠B) = 67.5°.
∵CE = CA,
∴∠CAE = ∠E = $\frac{1}{2}$∠ACB = 22.5°,
∴∠BAE = 180° - ∠B - ∠E = 112.5°,
∴∠DAE = ∠BAE - ∠BAD = 45°.
(2)
∵AB = AC,∠BAC = 120°,
∴∠B = ∠ACB = 30°.
∵BA = BD,
∴∠BAD = ∠BDA = $\frac{1}{2}$(180° - ∠B) = 75°,
∴∠DAC = ∠BAC - ∠BAD = 45°
∵CA = CE,
∴∠E = ∠CAE = $\frac{1}{2}$∠ACB = 15°,
∴∠DAE = ∠DAC + ∠CAE = 60°.
(3)∠DAE = $\frac{1}{2}$∠BAC.理由如下:
设∠CAE = x、∠BAD = y,
则∠B = 180° - 2y,∠E = ∠CAE = x,
∴∠BAE = 180° - ∠B - ∠E = 2y - x,
∴∠DAE = ∠BAE - ∠BAD = 2y - x - y = y - x,∠BAC = ∠BAE - ∠CAE = 2y - x - x = 2y - 2x,
∴∠DAE = $\frac{1}{2}$∠BAC.
14. (2024·兰州中考)如图,在$\triangle ABC$中,$AB= AC$,$\angle BAC= 130^\circ$,$DA \perp AC$,则$\angle ADB= $(
A.$100^\circ$
B.$115^\circ$
C.$130^\circ$
D.$145^\circ$
B
).A.$100^\circ$
B.$115^\circ$
C.$130^\circ$
D.$145^\circ$
答案：14.B [解析]在△ABC中,AB = AC,
∴∠B = ∠C;
∵∠BAC = 130°,
∴∠B = ∠C = $\frac{180^\circ - 130^\circ}{2}$ = 25°.
∵DA⊥AC,
∴∠DAC = 90°,
∴∠ADC = 90° - 25° = 65°,
∴∠ADB = 180° - ∠ADC = 180° - 65° = 115°.故选B.
∴∠B = ∠C;
∵∠BAC = 130°,
∴∠B = ∠C = $\frac{180^\circ - 130^\circ}{2}$ = 25°.
∵DA⊥AC,
∴∠DAC = 90°,
∴∠ADC = 90° - 25° = 65°,
∴∠ADB = 180° - ∠ADC = 180° - 65° = 115°.故选B.
15. (2024·重庆中考)如图,在$\triangle ABC$中,$AB= AC$,$\angle A= 36^\circ$,$BD平分\angle ABC交AC于点D$. 若$BC= 2$,则$AD$的长度为______.
2
答案：15.2 [解析]
∵AB = AC,
∴∠ABC = ∠C.
∵∠A + ∠ABC + ∠C = 180°,∠A = 36°,
∴∠ABC = ∠C = 72°.
∵BD平分∠ABC,
∴∠CBD = ∠ABD = 36°,
∴∠BDC = 180° - ∠C - ∠CBD = 180° - 72° - 36° = 72°,
∴∠BDC = ∠C,
∴BD = BC = 2,
∵∠A = 36°,∠ABD = 36°,
∴∠A = ∠ABD,
∴AD = BD = 2.
∵AB = AC,
∴∠ABC = ∠C.
∵∠A + ∠ABC + ∠C = 180°,∠A = 36°,
∴∠ABC = ∠C = 72°.
∵BD平分∠ABC,
∴∠CBD = ∠ABD = 36°,
∴∠BDC = 180° - ∠C - ∠CBD = 180° - 72° - 36° = 72°,
∴∠BDC = ∠C,
∴BD = BC = 2,
∵∠A = 36°,∠ABD = 36°,
∴∠A = ∠ABD,
∴AD = BD = 2.