a) Xét ΔAKB và ΔAKC có:
AB=AC(gt)
AK:cạnh chung
BK=CK(gt)
=> ΔAKB=ΔAKC(c.c.c)
=> \(\widehat{AKB}=\widehat{AKC}\)
Mà: \(\widehat{AKB}+\widehat{AKC}=180^o\)
=> \(\widehat{AKB}=\widehat{AKC}=90^o\)
=> \(AK\perp BC\)
b) Vì: \(EC\perp BC\left(gt\right)\)
Mad: \(AK\perp BC\left(cmt\right)\)
=> EC//AK