Question

In given figure ABC is a triangle in which AB=AC and D is any point in BC. Prove that ${AB}^2-{AD}^2=BD.CD$

Answer 1

Draw $AE\perp BC$

In $△AEB$ and $△AEC$, we have

$AB=AC$

$AE=AE$ [common]

and, $\angle b=\angle c$ [$because$ $AB=AC$]

$\therefore$ $△AEB\cong △AEC$

$\Rightarrow$ $BE=CE$

Since $△AED$ and $△ABE$ are right-angled triangles at E.

Therefore,

${AD}^2={AE}^2+{DE}^2$ and ${AB}^2={AE}^2+{BE}^2$

$\Rightarrow$ ${AB}^2-{AD}^2={BE}^2-{DE}^2$

$\Rightarrow$ ${AB}^2-{AD}^2=(BE+DE)(BE-DE)$

$\Rightarrow$ ${AB}^2-{AD}^2=(CE+DE)(BE-DE)$ [$\because BE=CE$]

$\Rightarrow$ ${AB}^2-{AD}^2=CD.BD$

${AB}^2-{AD}^2=BD.CD$ $\left[Henceproved\right]$