Question

In a right angled triangle ABC, $\angle B={90}^{\circ }$. If $BC$ is trisected by the point $D$ and $E$.

prove that $8{AE}^2=3{AC}^2+5{AD}^2$

Answer 1

In $\Delta ABCA{C}^2=A{B}^2+B{C}^2$...(1)

Multiply by 3 we get $3A{C}^2=3A{B}^2+B{C}^2$

$3A{C}^2=3A{B}^2+3\left(\frac{3}{2}BE{)}^2\right[BC=\frac{3}{2}BE]$

$3A{C}^2=3A{B}^2+\frac{27}{4}B{E}^2$ ....(2)

In $\Delta ABD$

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

Multiply by 5 we get $5A{D}^2=5A{B}^2+5B{D}^2(BD=\frac{1}{2}BE)$

$5AD=5A{B}^2+\frac{5}{4}B{E}^2$ .....(3)

on adding eq. (2) & (3) we get

$3A{C}^2+5A{D}^2=8A{B}^2+\frac{32}{4}B{E}^2$

$3A{C}^2+5A{D}^2=8(A{B}^2+B{E}^2)$

$[3A{C}^2+5A{D}^2=8A{E}^2]\left[from\Delta ABE\right]$

Henece Prove