a) Xét $\triangle$ABE và $\triangle$ACD có:
$\widehat{BAC}$ chung
AB = AC ( $\triangle$ABE cân)
AE = AD (gt)
Vậy $\triangle$ABE = $\triangle$ACD (c.g.c)
`=>` BE = CD ( 2 cạnh tương ứng)
b) Vì $\triangle$ABE = $\triangle$ACD (cmt)
`=>` $\widehat{ABE}$ = $\widehat{ACD}$
c) $\triangle$ABC cân `=>` $\widehat{ABC}$ = $\widehat{ACB}$
Mà $\widehat{ABE}$ = $\widehat{ACD}$ (cmt)
`=>` $\widehat{ABC}$ - $\widehat{ABE}$ = $\widehat{ACB}$ - $\widehat{ACD}$
`<=>` $\widehat{KBC}$ = $\widehat{KCB}$
`<=>` $\triangle$KBC cân tại K