$解：点C，D，E在一条直线上，理由：$
$连接CD，ED$
$在△ADC和△BDC中$
$\begin{cases}{AC=BC }\\{AD=BD} \\ {CD=CD} \end{cases}$
$∴△ADC≌△BDC(\mathrm {SSS})$
$∴∠ADC=∠BDC$
$在△ADE和△BDE中$
$\begin{cases}{AD=BD }\\{AE=BE} \\ {ED=ED} \end{cases}$
$∴△ADE≌△BDE(\mathrm {SSS})$
$∴∠ADE=∠BDE$
$∵∠ADC+∠BDC+∠ADE+∠BDE= 360°$
$∴2∠ADC+2∠ADE= 360°$
$∴∠ADC+∠ADE= 180°$
$∴点C，D，E在一条直线上$
