he thuc luong trong tam giac

S

sofia1997

VT=$\frac{1}{2}(2bc.cos A + 2ca.cos B +2ab.cos C)$
$=\frac{1}{2}(b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2)$
$=\frac{a^2+b^2+c^2}{2}$
VT=VP [TEX]\Leftrightarrow[/TEX] $b^2+ c^2=a^2$
[TEX]\Rightarrow[/TEX]tam giác ABC vuông
 
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