[Toán 8] Bài hình nâng cao

N

ngocbich74

Vì tg ABC đều \RightarrowAB=BC=AC=a

Ta có $S_{AMB}$+$S_{AMC}$+$S_{BMC}$=$S_{ABC}$

Hay $\dfrac{KMa+MFa+MIa}{2}$=$\dfrac{AHa}{2}$

\RightarrowKMa+MFa+MIa=AHa

Hay a(KM+MF+MI)=AHa\RightarrowKM+MF+MI=AH
 
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