$证明:延长AE到点F使得EF=AE,连接DF$
$在△ABE和△FDE中$
${{\begin{cases} {{BE=DE}} \\ {∠AEB=∠FED} \\ {AE=FE} \end{cases}}}$
$∴△ABE≌△FDE(SAS)$
$∴AB=FD,∴∠ABE=∠FDE$
$∴FD=DC,∠FDE+∠ADE=∠ABE+∠BAD=∠ADC$
$在△ADF和△ADC中$
${{\begin{cases} {{AD=AD}} \\ {∠ADF=∠ADC} \\ {DF=DC} \end{cases}}}$
$∴△ADF≌△ADC(SAS)$
$∴AF=AC$
$∴AC=AE+EF=2AE$