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a)Kẻ $CH ; MK ; BI \bot AB$

Ta có: $KM=\dfrac{BI+CH}{2}$

$\rightarrow 2S_{ADM}=KM.AB=\dfrac{(BI+CH)AB}{2}$

$\rightarrow 2S_{CND}+2S_{BNA}=\dfrac{CH.AB}{2}+\dfrac{BI.AB}
{2}=\dfrac{(BI+CH)AB}{2}$

$\rightarrow S_{ADM}=S_{NCD}+S_{BNA} \rightarrow (dpcm)$

Câu b hỏi gì bạn
 
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