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20. (8分)如图，在△ABC中，AD⊥BC，且AD = BD，点E是线段AD上一点，且BE = AC.
(1)求证：△ACD≌△BED.
(2)若∠C = 78°，求∠ABE的度数.
(1)求证：△ACD≌△BED.
(2)若∠C = 78°，求∠ABE的度数.
答案：
20.
(1)证明：$\because AD \perp BC$，$\therefore \angle ADC = \angle BDE = 90^{\circ}$.
在$Rt \triangle ACD$和$Rt \triangle BED$中，$\begin{cases} AC = BE, \\ AD = BD, \end{cases}$
$\therefore Rt \triangle ACD \cong Rt \triangle BED (HL)$.
(2)解：由
(1)得，$\triangle ACD \cong \triangle BED$，$\therefore \angle DAC = \angle DBE$.
$\because \angle C = 78^{\circ}$，$\therefore \angle DBE = \angle DAC = 90^{\circ} - \angle C = 12^{\circ}$.
$\because AD = BD$，$AD \perp BC$，
$\therefore \triangle ABD$是等腰直角三角形.$\therefore \angle ABD = 45^{\circ}$.
$\therefore \angle ABE = \angle ABD - \angle DBE = 45^{\circ} - 12^{\circ} = 33^{\circ}$.
(1)证明：$\because AD \perp BC$，$\therefore \angle ADC = \angle BDE = 90^{\circ}$.
在$Rt \triangle ACD$和$Rt \triangle BED$中，$\begin{cases} AC = BE, \\ AD = BD, \end{cases}$
$\therefore Rt \triangle ACD \cong Rt \triangle BED (HL)$.
(2)解：由
(1)得，$\triangle ACD \cong \triangle BED$，$\therefore \angle DAC = \angle DBE$.
$\because \angle C = 78^{\circ}$，$\therefore \angle DBE = \angle DAC = 90^{\circ} - \angle C = 12^{\circ}$.
$\because AD = BD$，$AD \perp BC$，
$\therefore \triangle ABD$是等腰直角三角形.$\therefore \angle ABD = 45^{\circ}$.
$\therefore \angle ABE = \angle ABD - \angle DBE = 45^{\circ} - 12^{\circ} = 33^{\circ}$.
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