登录
7. (2024重庆中考)如图,在$\triangle ABC$中,$AB = AC$,$∠A = 36^{\circ}$,$BD平分∠ABC交AC于点D$。若$BC = 2$,则$AD$的长为______。
答案
2
8. (教材变式)如图,在$\triangle ABC$中,$AB = AC$,$D为AC$上一点,连接$BD$,作$∠CDE = ∠ABD$,交
$BC的延长线交于点E$。求证：$BD = DE$。
$BC的延长线交于点E$。求证：$BD = DE$。
答案
证明：$\because AB = AC$，
$\therefore \angle ABC = \angle ACB$，
$\because \angle ABD = \angle CDE$，
$\therefore \angle ABC - \angle ABD = \angle ACB - \angle CDE$，
即 $\angle DBC = \angle E$，
$\therefore BD = DE$。
$\therefore \angle ABC = \angle ACB$，
$\because \angle ABD = \angle CDE$，
$\therefore \angle ABC - \angle ABD = \angle ACB - \angle CDE$，
即 $\angle DBC = \angle E$，
$\therefore BD = DE$。
9. (2025原创题)如图,$BF平分∠ABC$,$FA\perp AB$,$FC\perp BC$,$AD\perp BC于点D$,交$BF于点E$。求证：$AE = FC$。
答案
证明：$\because BF$ 平分 $\angle ABC$，$FA \perp AB$，
$FC \perp BC$，
$\therefore FA = FC$。
$\because FA \perp AB$，$AD \perp BC$，
$\therefore \angle AFB + \angle ABF = 90^{\circ}$，
$\angle DEB + \angle DBE = 90^{\circ}$。
$\because \angle ABF = \angle DBE$，
$\therefore \angle AFB = \angle DEB$。
$\because \angle AEF = \angle DEB$，
$\therefore \angle AFB = \angle AEF$，
$\therefore AE = AF$，
$\therefore AE = FC$。
$FC \perp BC$，
$\therefore FA = FC$。
$\because FA \perp AB$，$AD \perp BC$，
$\therefore \angle AFB + \angle ABF = 90^{\circ}$，
$\angle DEB + \angle DBE = 90^{\circ}$。
$\because \angle ABF = \angle DBE$，
$\therefore \angle AFB = \angle DEB$。
$\because \angle AEF = \angle DEB$，
$\therefore \angle AFB = \angle AEF$，
$\therefore AE = AF$，
$\therefore AE = FC$。
10. (2025南通)如图,在$\triangle ABC$中,$∠B = 30^{\circ}$,$∠ACB = 105^{\circ}$,$AD是\triangle ABC$的角平分线。求证：$AB = AC + BD$。
答案
证明：在 $AB$ 上截取 $AE = AC$，连接 $DE$。
$\because AD$ 平分 $\angle BAC$，
$\therefore \angle BAD = \angle DAC$。
又 $\because AE = AC$，$AD = AD$，
$\therefore \triangle ADE \cong \triangle ADC(SAS)$，
$\therefore \angle AED = \angle ACD = 105^{\circ}$，
$\therefore \angle BED = 75^{\circ}$。
$\because \angle B = 30^{\circ}$，
$\therefore \angle BDE = 75^{\circ} = \angle BED$，
$\therefore BD = BE$，
$\therefore AB = AE + BE = AC + BD$。
$\because AD$ 平分 $\angle BAC$，
$\therefore \angle BAD = \angle DAC$。
又 $\because AE = AC$，$AD = AD$，
$\therefore \triangle ADE \cong \triangle ADC(SAS)$，
$\therefore \angle AED = \angle ACD = 105^{\circ}$，
$\therefore \angle BED = 75^{\circ}$。
$\because \angle B = 30^{\circ}$，
$\therefore \angle BDE = 75^{\circ} = \angle BED$，
$\therefore BD = BE$，
$\therefore AB = AE + BE = AC + BD$。
11. (2025武汉二中)如图,$AD为\triangle ABC$的角平分线,$CE\perp AD交AD的延长线于点E$,$∠BAD = 2∠DCE$。
(1)求证：$\triangle ABD$为等腰三角形；
(2)求证：$AD + AC = 2AE$。
(1)求证：$\triangle ABD$为等腰三角形；
(2)求证：$AD + AC = 2AE$。
答案
证明：(1)设 $\angle DCE = x$，
$\therefore \angle BAD = 2\angle DCE = 2x$。
$\because CE \perp AE$，
$\therefore \angle ADB = \angle CDE = 90^{\circ} - x$，
$\therefore \angle B = 180^{\circ} - \angle BAD - \angle ADB$
$= 90^{\circ} - x$，
$\therefore \angle B = \angle ADB$，
$\therefore AB = AD$，
$\therefore \triangle ABD$ 为等腰三角形；
(2)过点 $C$ 作 $CF // AB$ 交 $AE$ 的延长线于点 $F$，
$\therefore \angle FCD = \angle B$，$\angle F = \angle BAD$。
$\because AD$ 平分 $\angle BAC$，
$\angle B = \angle ADB$，
$\therefore \angle BAD = \angle CAF$，
$\angle FCD = \angle ADB = \angle FDC$，
$\therefore \angle F = \angle CAF$，$CF = FD$，
$\therefore AC = CF = DF$，
$\because AD + AC = AD + DF = AF$。
$\because AC = CF$，$CE \perp AF$，
$\therefore AE = EF$，
$\therefore AD + AC = AF = 2AE$。
注：也可延长 $AE$ 至点 $F$，使 $EF =$ $AE$ 来证明。
$\therefore \angle BAD = 2\angle DCE = 2x$。
$\because CE \perp AE$，
$\therefore \angle ADB = \angle CDE = 90^{\circ} - x$，
$\therefore \angle B = 180^{\circ} - \angle BAD - \angle ADB$
$= 90^{\circ} - x$，
$\therefore \angle B = \angle ADB$，
$\therefore AB = AD$，
$\therefore \triangle ABD$ 为等腰三角形；
(2)过点 $C$ 作 $CF // AB$ 交 $AE$ 的延长线于点 $F$，
$\therefore \angle FCD = \angle B$，$\angle F = \angle BAD$。
$\because AD$ 平分 $\angle BAC$，
$\angle B = \angle ADB$，
$\therefore \angle BAD = \angle CAF$，
$\angle FCD = \angle ADB = \angle FDC$，
$\therefore \angle F = \angle CAF$，$CF = FD$，
$\therefore AC = CF = DF$，
$\because AD + AC = AD + DF = AF$。
$\because AC = CF$，$CE \perp AF$，
$\therefore AE = EF$，
$\therefore AD + AC = AF = 2AE$。
注：也可延长 $AE$ 至点 $F$，使 $EF =$ $AE$ 来证明。
