$\text{a.Xét ΔBKA và ΔBHD có :}$
$\widehat{BKA}=\widehat{BHD}=90^o$
$\text{BD : chung}$
$\widehat{KBA}=\widehat{HBD}$
$\text{⇒ΔBKA=ΔBHD(ch-gn)}$
$\text{⇒BK=BH ( 2 cạnh tương ứng )}$
$\text{b.Ta có :}$
$\text{AB=KB+KA}$
$\text{BE=BH+EH}$
$\text{Mà AB=BE(gt);BK=BH(ΔBKA=ΔBHD)}$
$\text{⇒KA=EH}$
$\text{Xét ΔAKD và ΔEHD có :}$
$\widehat{AKD}=\widehat{EHD}=90^o$
$\text{KA=EH(cmt)}$
$\text{KD=KH(ΔBKA=ΔBHD)}$
$\text{⇒ΔAKD=ΔEHD(c.g.c)}$