a) Ta có:
\(\left[ \begin{array}{l}\widehat{A1}+\widehat{A2}=90^{0}\\\widehat{A2}+\widehat{C2}=90^{0}\end{array} \right.\)
$⇒\widehat{A1}=\widehat{C1}$
Xét $ΔABD$ và $ΔCAE$, có:
\(\left[ \begin{array}{l}\widehat{ADB}=\widehat{CEA}=90^{0}$\\\widehat{A1}=\widehat{C1}(CMT)\\AB=CA\end{array} \right.\)
$⇒ΔABD=ΔCAE$ $(CH-GN)$
$⇒ AD=CE$ (2 cạnh tương ứng)