If ABC is an isosceles triangle and D is a point of BC such that AD ⊥ BC, then
(a) $\mathrm{AB}^{2}-\mathrm{AD}^{2}=\mathrm{BD} . \mathrm{DC}$
(b) $A B^{2}-A D^{2}=B D^{2}-D C^{2}$
(c) $A B^{2}+A D^{2}=B D . D C$
(d) $A B^{2}+A D^{2}=B D^{2}-D C^{2}$
Given: $\triangle \mathrm{ABC}$ is an isosceles triangle, $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\mathrm{AD} \perp \mathrm{BC}$
We know that in an isosceles triangle the perpendicular from the vertex bisects the base.
$\therefore \mathrm{BD}=\mathrm{DC}$
Applying Pythagoras theorem in ΔABD
$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$
$\Rightarrow \mathrm{AB}^{2}-\mathrm{AD}^{2}=\mathrm{BD}^{2}$
$\Rightarrow \mathrm{AB}^{2}-\mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{BD}$
Since $\mathrm{BD}=\mathrm{DC}$
$\Rightarrow \mathrm{AB}^{2}-\mathrm{AD}^{2}=\mathrm{BD} \times \mathrm{DC}$
Hence correct answer is $(a)$.