A,Ta có $\triangle CMD\sim \triangle CAB$
$\Rightarrow CD.CA=CM.CB=\frac{CB^{2}}{2}$
b,Ta có $AH^{2}=BH.CH$
$AH=9,6$
$S_{ABC}=\frac{1}{2}AH.BC=96$
Ta có CM=10=BM
$AB^{2}=BH.BC$
$\Rightarrow AB=12$
Tương tự AC=16
Ta có $\frac{CM}{CA}=\frac{MD}{AB}$
$\Rightarrow MD=7,5$
$\Rightarrow S_{CMD}=37,5$
c,Ta có CD=12,$MD^{2}=DK.DC\Rightarrow DK=4,5$
$MD^{2}=DK^{2}+MK^{2} \Rightarrow MK=6$
$\Rightarrow S_{MDK}=\frac{1}{2}MK.DK=13,5$