a,
$\hat{CAE}= \hat{BAC}- \hat{BAD}$
$\hat{ABD}= \hat{BAC}- \hat{BAD}$
=> $\hat{CAE}= \hat{ABD}$
$\Delta$ ADB và $\Delta$ CEA có:
$\hat{ADB}= \hat{CEA}= 90^o$
$\hat{ABD}= \hat{CAE}$
AB= AC
=> $\Delta$ ADB= $\Delta$ CEA (ch.gn) (*)
b,
(*)=> BD= AE, AD= EC
Có AE= AD+ DE
=> BD= CE+ DE
<=> BD-CE= DE