hình 9

C

congchuaanhsang

Xét $\Delta$HCD và $\Delta$ABM có:
$\hat{BAM}$=$\hat{CHD}$ (=$90^0$) ; $\hat{DCH}$=$\hat{ABM}$ (cùng phụ với $\hat{AMB}$)
\Rightarrow$\Delta$HCD $\sim$ $\Delta$ABM (g.g)
\Rightarrow$\dfrac{AB}{AM}$=$\dfrac{HC}{HD}$=2 \Rightarrow HC=2HD
Đặt HD=x\RightarrowHC=2x. Ta có:
$DH^2$=HM.HC\Leftrightarrow$x^2$=HM.2x \Leftrightarrow HM=0,5x
\RightarrowMC=0,5x+2x=2,5x=AM\RightarrowAH=2,5x+0,5x=3x=3HD
Vậy AH=3HD
 
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