Giải thích các bước giải:

1.Ta có : $HM\perp BC\to\widehat{CMH}=\widehat{CAB}=90^o$ 

$\to \Delta CMH\sim\Delta CAB(g.g)$

2.Ta có : $BC^2=AB^2+AC^2=100\to BC=10$

Từ câu 1 $\to \dfrac{CH}{CB}=\dfrac{HM}{AB}$
$\to HM=\dfrac{CH.AB}{CB}=3$

3.a.Ta có : $\widehat{HMC}=\widehat{HAD}=90^o,\widehat{AHD}=\widehat{MHC}$
$\to \Delta AHD\sim\Delta MHC(g.g)$
$\to \dfrac{HA}{HM}=\dfrac{HD}{HC}$

$\to \dfrac{HA}{HD}=\dfrac{HM}{HC}$

Mà $\widehat{AHM}=\widehat{DHC}$
$\to \Delta AHM\sim\Delta DHC(c.g.c)\to \widehat{MAH}=\widehat{HDC}$

Ta có $CA\perp AB, DM\perp BC\to H$ là trực tâm $\Delta ABC\to BH\perp CD$

$\to \widehat{BMH}=\widehat{HID}=90^o$

Mà $\widehat{BHM}=\widehat{DHI}\to\Delta BHM\sim\Delta DHI(g.g)$
$\to \widehat{HBM}=\widehat{HDI}=\widehat{HDC}=\widehat{HAM}$

b.Ta có : $\widehat{DMC}=\widehat{BMH}=90^o$

Mà $\widehat{MBH}=\widehat{HDC}=\widehat{MDC}$ (câu 3a)

$\to \Delta BHM\sim\Delta DCM(g.g)$
$\to \dfrac{MB}{MD}=\dfrac{MH}{MC}\to MH.MD=MB.MC\le\dfrac{(MB+MC)^2}4=\dfrac{BC^2}4$

Dấu = xảy ra khi $MB=MC\to M$ là trung điểm BC