Question

In the fifure $BL$ & $CM$ are the medians of $\Delta ABC$, right angled at $A$. Prove that $4(B{L}^2+C{M}^2)=5B{C}^2$.

Answer 1

Given:

$\Delta ABC$ is a right angled at $A$, i.e., $\angle A=90{}^{\circ }$, where $BL$ and $CM$ are the medians

To prove:

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

Proof:

Since, $BL$ is the median,

$AL=CL=\frac{1}{2}AC$ …… (1)

Similarly, $CM$ is the median,

$AM=MB=\frac{1}{2}AB$ …… (2)

We know that, by Pythagoras theorem,

${\left(Hypotenuse\right)}^2={\left(Height\right)}^2+{\left(Base\right)}^2$

Therefore,

In $\Delta BAC$,

$B{C}^2=A{B}^2+A{C}^2$

In $\Delta BAL$,

$B{L}^2=A{B}^2+A{L}^2$

$B{L}^2=A{B}^2+{\left(\frac{AC}{2}\right)}^2$

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

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

In $\Delta MAC$,

$C{M}^2=A{M}^2+A{C}^2$

$C{M}^2={\left(\frac{AB}{2}\right)}^2+A{C}^2$

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

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

Now,

$B{C}^2=A{B}^2+A{C}^2$ …… (3)

$4B{L}^2=4A{B}^2+A{C}^2$ …… (4)

$4C{M}^2=A{B}^2+4A{C}^2$ …… (5)

Add equations (4) and (5).

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

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

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

Hence, proved.