Giải thích các bước giải:
$a)Xét$ $ΔADC$ ,$ΔDAE$ $có:$
$Chung$ $cạnh$ $AD$
$\widehat{CDA}=\widehat{EAD}$ $(2$ $góc$ $so$ $le$ $trong)$
$DC=AE$ $(gt)$
$⇒ΔADC=ΔDAE(cgc)$
$b)Có:$
$\widehat{B}=\widehat{HAC}(do$ $cùng$ $phụ$ $với$ $\widehat{BAH})$
$Lại có$ $\widehat{DAH}=\widehat{C}$
$⇒\widehat{DAC}+\widehat{DAH}=\widehat{DAC}+\widehat{C}$
$⇒\widehat{HAC}=\widehat{ADB}$
$⇒\widehat{ABD}=\widehat{ADB}$
$⇒ΔABD$ $cân$ $tại$ $A.$