a,
$\Delta$ ABC và $\Delta$ ADE có:
$\widehat{BAC}= \widehat{DAE}= 90^o$
AD= AB
AE= AC
=> $\Delta$ ABC= $\Delta$ ADE (c.g.c) (*)
b,
$\widehat{BAH}= 90^o- \widehat{ABC}$
$\widehat{ACH}= 90^o- \widehat{ABC}$
=> $\widehat{BAH}= \widehat{ACH}$
c,
(*)=> $\widehat{AED}= \widehat{ACB}$
Mà $\widehat{EAK}= \widehat{BAH}$
Theo b, $\widehat{BAH}= \widehat{ACB}$
=> $\widehat{AED}= \widehat{EAK}$
=> $\Delta$ KAE cân A
=> KA= KE
CMTT, KA= KD
=> KD= KE => K trung điểm DE