In $\Delta ABC,~AB~=~15$ cm and $AC~=~20$ cm. Medians BE and CF intersect each other at right angle. Find BC.
Solution
In $\Delta BGC$, it is given $\angle G~=~90{}^\circ ~$
Therefore,$~B{{G}^{2~}}+~C{{G}^{2~}}=~B{{C}^{2}}$
Now $\Delta BFG$ is at $\angle BGF~=~90{}^\circ $ and $\Delta CGE$ is at $\angle CGE~=~90{}^\circ $
$B{{F}^{2~}}~F{{G}^{2~}}+~C{{E}^{2~}}~G{{E}^{2~}}=~B{{C}^{2}}$
Now $BF~=~{\scriptscriptstyle 1\!/\!{ }_2}~AB$ and $CE~=~{\scriptscriptstyle 1\!/\!{ }_2}~AC$ and $FG2~+~GE2~=~F{{E}^{2~}}=~{{\left( \frac{BC}{2} \right)}^{2}}$
Therefore,
$\frac{A{{B}^{2}}}{4}+\frac{A{{C}^{2}}}{4}\frac{B{{C}^{2}}}{4}=~B{{C}^{2}}$
$A{{B}^{2~}}+~A{{C}^{2~}}=~5B{{C}^{2}}$
${{15}^{2~}}+~{{20}^{2~}}=~5B{{C}^{2}}$
$225~+~400~=~5B{{C}^{2}}$
$B{{C}^{2~}}=\frac{625}{5}=~125$
$BC~=~5\sqrt{5}cm$