Giải thích các bước giải:
Kẻ $MD\perp BH$
$\to MD//AC$ vì $BH\perp AC$
Mà $MF\perp AC\to MF//BH$
Xét $\Delta MDH,\Delta HMF$ có:
$\widehat{DHM}=\widehat{HMF}$ vì $DH//MF$
Chung $MH$
$\widehat{DMH}=\widehat{MHF}$ vì $MD//HF$
$\to\Delta DMH\sim\Delta FHM(g.c.g)$
$\to MF=DH$
Xét $\Delta BME,\Delta BMD$ có:
$\widehat{BEM}=\widehat{BDM}=90^o$
Chung $BM$
$\widehat{EBM}=\widehat{ABC}=\widehat{ACB}=\widehat{DMB}$
$\to \Delta EBM=\Delta DMB$(cạnh huyền-góc nhọn)
$\to ME=BD$
$\to BH=BD+DH=ME+MF$