Question

In the given figure,$AD\perp BC$ and $BD=3CD$ . Prove that $2A{B}^2=2A{C}^2+B{C}^2$

Answer 1

$In△ABD$

$A{B}^2=B{D}^2+A{D}^2-----\left(1\right)$

$In△AOC$

$A{C}^2=A{D}^2+D{C}^2-----\left(2\right)$

Subtract $\left(1\right)$ from $\left(2\right)$

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

$A{C}^2–A{B}^2=C{D}^2–B{D}^2-----\left(3\right)$

Given $BD=3CD$

And from figure $BD+DC=BC$

$⟹3CD+DC=BC$

$⟹CD=\frac{BC}{4}$

$⟹BD=\frac{3BC}{4}$

Put in $\left(3\right)$, we get

$A{C}^2–A{B}^2=\frac{B{C}^2}{16}-\frac{9B{C}^2}{16}=-\frac{8BC}{16}=-\frac{B{C}^2}{2}$

$⟹2A{C}^2–2A{B}^2=-B{C}^2$

$⟹2A{B}^2=2A{C}^2+B{C}^2$