Lời giải:
Ta có:
$AC//BD\quad (\perp AB)$
$\Rightarrow \triangle AHC\backsim \triangle DHB$
$\Rightarrow \dfrac{HA}{HD} = \dfrac{HC}{HB} = \dfrac{AC}{BD} = \dfrac{2AN}{2MD} = \dfrac{AN}{MD}$
Xét $\triangle AHN$ và $\triangle DHM$ có:
$\begin{cases}\dfrac{HA}{HD} = \dfrac{AN}{MD}\quad (cmt)\\\widehat{HAN} = \widehat{HDM}\quad \text{(so le trong)}\end{cases}$
Do đó $\triangle AHN\backsim \triangle DHM\ (c.g.c)$
$\Rightarrow \widehat{AHN} = \widehat{DHM}$
mà $A,H,D$ thẳng hàng
nên $M,H,N$ thẳng hàng
