Question

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

Answer 1

$\because ACB$ is isos. $\Delta$,

$AO\perp BC\alpha BO=CO$ :

$T.P.:A{D}^2=A{C}^2+BD\times CD$

Proof : $\begin{array}{c}A{D}^2=A{O}^2+O{D}^2\\ A{B}^2=A{O}^2+B{O}^2\end{array}\}\begin{array}{c}Pythogearus....\left(i\right)\\ theorem....\left(ii\right)\end{array}$

Subtract $\left(ii\right)$ from $\left(i\right)$

$A{D}^2-A{B}^2=O{D}^2-B{O}^2$

$A{D}^2-A{B}^2=(OD+BO)(OD-BO)[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$A{D}^2-A{B}^2=\left(BD\right)(OD-OC)[\because BO=CO]$

$A{D}^2-A{B}^2=\left(BD\right)\left(CD\right)[\because OD-OC=CD]$

$A{D}^2=A{B}^2+BD\times CD$

$A{D}^2=A{C}^2+BD\times CD[\because AB=AC\&A{B}^2=A{C}^2]$

Hence , proved