$\text{a) Xét ΔBDA và ΔBEA có:}$
$\text{BD chung}$
$\text{$\widehat{ABD}$ = $\widehat{EBD}$ (BD là p/g $\widehat{B}$)}$
$\text{BA = BE (gt)}$
$\text{⇒ ΔBDA = ΔBEA (c.g.c) (1)}$
$\text{⇒ $\widehat{BAD}$ = $\widehat{BED}$ (2 góc t/ứ)}$
$\text{mà $\widehat{BAD}$ = $90^{o}$ (ΔABC vuông tại A)}$
$\text{⇒ $\widehat{BED}$ = $90^{o}$}$
$\text{⇒ DE ⊥ BC (DHNB)}$
$\text{b) Có: $\widehat{BAD}$ = $\widehat{BED}$ (cmt)}$
$\text{mà $\widehat{BAD}$ + $\widehat{KAD}$ = $180^{o}$ (kề bù)}$
$\text{$\widehat{BED}$ + $\widehat{CED}$ = $180^{o}$ (kề bù)}$
$\text{⇒ $\widehat{KAD}$ = $\widehat{CED}$}$
$\text{Xét ΔADK và ΔEDC có:}$
$\text{$\widehat{KAD}$ = $\widehat{CED}$ (cmt)}$
$\text{từ (1) ⇒ AD = ED (2 cạnh t/ứ)}$
$\text{$\widehat{KDA}$ = $\widehat{CDE}$ (đối đỉnh)}$
$\text{⇒ ΔADK = ΔEDC (g.c.g)}$$
$\text{⇒ KA = CE (2 cạnh t/ứ)}$
