$證明：過點(diǎn)C作CF⊥AD交AD的延長(zhǎng)線于點(diǎn)F$
$∵AC平分∠BAD，CE⊥AB，CF⊥AD$
$∴CE=CF$
$∵∠B+∠ADC=180°，∠ADC+∠CDF=180°$
$∴∠B=∠CDF$
$在△BEC和△DFC中：$
$\begin{cases}{∠B=∠CDF}\\{∠BEC=∠DFC}\\{CE=CF}\end{cases}$
$∴△BEC≌△DFC(AAS)$
$∴BE=DF，CE=CF$
$在Rt△ACE和Rt△ACF中：$
$\begin{cases}{CE=CF} \\ {AC=AC}\end{cases}$
$∴Rt△ACE≌Rt△ACF(HL)$
$∴AE=AF$
$∵AF=AD+DF，DF=BE$
$∴AF=AD +BE\ $
$∴AE=AD+BE$
