Giải thích các bước giải:
a.Ta có: $MH\perp AB, AB\perp AC\to MH//AC$
$\to \widehat{BMH}=\widehat{BCA}$ (đồng vị)
Ta có: $MK\perp AC, AB\perp AC\to MK//AB$
$\to \widehat{KMC}=\widehat{HBM}$ (đồng vị)
b.Ta có:
$\widehat{HMK}=180^o-(\widehat{HMB}+\widehat{KMC})$
$\to \widehat{HMK}=180^o-(\widehat{ACB}+\widehat{HBM})$
$\to \widehat{HMK}=180^o-(\widehat{ACB}+\widehat{ABC})$
$\to \widehat{HMK}=180^o-(180^o-\widehat{BAC})$
$\to \widehat{HMK}=180^o-(180^o-90^o)$
$\to \widehat{HMK}=90^o$
c.Theo câu a $\to HM//AC, MK//AB$