a,
Có $\widehat{AMB}=\widehat{CMD}$ (đối đỉnh)
Mà $\widehat{CMD}=\widehat{CDH}=90^o-\widehat{HDM}$
$\to \widehat{AMB}=\widehat{CDH}$
Mặt khác $\widehat{BAM}=\widehat{CHD}=90^o$
Vậy $\Delta BAM\backsim\Delta CHD$ (g.g)
b,
$AM=\dfrac{AC}{2}=\dfrac{AB}{2}=4(cm)$
$\Delta ABM$ vuông tại $A$ có:
$\tan\widehat{AMB}=\dfrac{AB}{AM}=\dfrac{8}{4}=2$
$\to \tan\widehat{DMC}=2$
$\to \cot\widehat{DMC}=\dfrac{1}{2}$
$\dfrac{1}{\sin^2\widehat{DMC}}=1+\cot^2\widehat{DMC}$
$\to \sin\widehat{DMC}=\dfrac{2\sqrt5}{5}$
$\Delta DMC$ vuông tại $D$ có:
$CD=MC.\sin\widehat{DMC}=4.\dfrac{2\sqrt5}{5}= \dfrac{8\sqrt5}{5}(cm)$
$DH\bot MC$ nên ta có:
$DC^2=HC.MC$
$\to HC=3,2(cm)$
$\to MH=MC-CH=0,8(cm)$
$\to DH=\sqrt{MH.HC}=1,6(cm)$
