Question

In the given figure, $ABC$ is a triangle right angled at $B.D$ and $E$ are points on $BC$ trisect it.Prove that $8{AE}^2=3{AC}^2+5{AD}^2$

Answer 1

Given right angle $\Delta ABC$, right-angled at $B$.

$D$ and $E$ are points of trisection of the side $BC$

Let $BD=DE=EC=k$

Hence we get $BE=2k$ and $BC=3k$

In $△ABD$, by Pythagoras theorem, we get

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

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

In $△ABE$ we get

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

Hence $A{E}^2=A{B}^2+{\left(2k\right)}^2$

$=A{B}^2+4k^2$ and

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

$=AB+{\left(3k\right)}^2$

$A{C}^2=A{B}^2+9k^2$

Consider, $3A{C}^2+5A{D}^2=3(A{B}^2+9k^2)+5(A{B}^2+4k^2)$

$=8A{B}^2+32k^2$

$=8(A{B}^2+4k^2)$

$\therefore 3A{C}^2+5A{D}^2=8A{E}^2$