Question

In $\Delta$ABC, $\angle$BAC$={90}^o$, seg BL and seg CM are medians of $\Delta$ABC. Then prove that: $4(B{L}^2+C{M}^2)=5B{C}^2$.

Answer 1

Given, $\Delta ABC$ is right-angled at $A$ i.e. $\angle A={90}^0$. where $BL,CM$ are medians.

Since $BL$ is median, $AL=CL=\frac{1}{2}AC$

Similary, $CM$ is median. $AM=MB=\frac{1}{2}AB$ ------(2)

Apply pythagoras theoram in $\Delta BAC$,

$B{C}^2=A{B}^2+A{C}^2$ ........(i)

In $\Delta BAL,B{L}^2=A{B}^2+A{L}^2\Rightarrow B{L}^2=A{B}^2+\frac{A{C}^2}{4}$

$\Rightarrow 4B{L}^2=4A{B}^2+A{C}^2$ ..........(ii)

In $\Delta MAC,C{M}^2=A{M}^2+A{C}^2\Rightarrow C{M}^2=\frac{A{B}^2}{4}+A{C}^2$

$\Rightarrow 4C{M}^2=A{B}^2+4A{C}^2$ .........(iii)

Adding (ii) and (iii)

$\Rightarrow 4B{L}^2+4C{M}^2=4A{B}^2+A{C}^2+A{B}^2+4A{C}^2$

$\Rightarrow 4(B{L}^2+C{M}^2)=5B{C}^2$

$\therefore$ Hence proved.