In Figure 1, BL and CM are medians of a right-angled at A. Prove that

In Figure 1, BL and CM are medians of a $\triangle ABC$ right-angled at A. Prove that $4\left ( BL^{2}+CM^{2} \right )= 5\, BC^{2}$

Answers (1)

Given in $\triangle ABC\rightarrow \angle A= 90^{\circ};B\angle$ and CM are medians
To Prove $\rightarrow 4\left ( BL^{2}+CM^{2} \right )= 5BC^{2}$

$AL= \left ( \frac{AC}{2} \right )---(i)\left ( \therefore AL= LC \right )$
$AM= \left ( \frac{AB}{2} \right )---(ii)\left ( \therefore AM= BM \right )$
Now in $\triangle ABC,\triangle BAL\, and\; \triangle MBC$
$BC^{2}= AB^{2}+AC^{2}---(iii)$ (using pythagarous in $\triangle ABC$ )
$CM^{2}= AC^{2}+BM^{2}---(iv)$ (uisng pythagarous in $\triangle MAC$ )
$BL^{2}= AB^{2}+AL^{2}---(v)$ (using pythagarous in $\triangle BAL$ )
Adding equation (iv) and (v)
$CM^{2}+BL^{2}= AB^{2}+AC^{2}+AL^{2}+AM^{2}$
$CM^{2}+BL^{2}= BC^{2}+AL^{2}+AM^{2}$ (using (iii))
$CM^{2}+BL^{2}= BC^{2}+\frac{AC^{2}}{4}+\frac{AB^{2}}{4}$ (using (i) and (ii))
$CM^{2}+BL^{2}= \frac{4BC^{2}+AC^{2}+AB^{2}}{4}$
$\Rightarrow 4\left ( CM^{2}+BL^{2} \right )= 5BC^{2}$ (using (iii))
Hence proved