Question

In an isosceles triangle, $ABC,AB=AC$ and $D$ is a point on $BC$ produced. Prove that
${AD}^2={AC}^2+BD.CD$

Answer 1

Let AX be a line perpendicular to BC and as perpendicular from a vertex common to equal side is also a median therefore, $BX=CX=\frac{1}{2}\left(BC\right)$

Now, By Pythagoras theorem in $\Delta ADX$ and $\Delta ACX$,

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

and $A{C}^2=A{X}^2+C{X}^2$

$\Rightarrow A{D}^2-A{C}^2=X{D}^2-C{X}^2$
$=(BD-BX{)}^2-C{X}^2$
$=B{D}^2+B{X}^2-2BD.BX-C{X}^2(BX=CX)$

$=BD(BD-2BX)$

$=BD(BD-BC)$

$\Rightarrow A{D}^2-A{C}^2=BD\cdot CD$

$\therefore A{D}^2=A{C}^2+BD\cdot CD$