Giải thích các bước giải:
a.Xét $\Delta AHC,\Delta DHB$ có:
$\widehat{AHC}=\widehat{BHD}$
$\widehat{HAC}=\widehat{HDB}$ vì $BD//AC$
$\to\Delta AHC\sim\Delta DHB(g.g)$
b.Xét $\Delta ABC,\Delta ABD$ có:
$\widehat{BAC}=\widehat{ABD}(=90^o)$
$\widehat{BAD}=90^o-\widehat{HAC}=\widehat{ACB}$
$\to\Delta ABD\sim\Delta CAB(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{BD}{AB}$
$\to AB^2=AC.BD$
c.Ta có $BD//AC$
$\to \dfrac{HB}{HC}=\dfrac{BD}{AC}=\dfrac{2BM}{2CN}=\dfrac{BM}{CN}$
Mà $\widehat{HBM}=\widehat{HCN}$
$\to\Delta HBM\sim\Delta HCN(c.g.c)$
$\to \widehat{BHM}=\widehat{CHN}$
$\to M, H, N$ thẳng hàng
