violympic 9

A

angleofdarkness

Goi O là giao điểm 2 đường trung tuyến đó.

Có $BO^2+OE^2=BE^2=\dfrac{AB^2}{4}=9 \\ CO^2+OD^2=CD^2=\dfrac{AC^2}{4}=16 $

Mà $OE=\dfrac{OC}{2}$ \Rightarrow $OE^2=\dfrac{OC^2}{4}$

$OD=\dfrac{BO}{2}$ \Rightarrow $OD^2=\dfrac{BO^2}{4}$

\Rightarrow $BO^2+OE^2+CO^2+OD^2 =\dfrac{5}{4}.(BO^2+OE^2) = 9+16 =25$

\Rightarrow $BC^2=BO^2+OE^2= 20$ \Rightarrow $BC=2\sqrt{5}$
 
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