Question

In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to the side AC. Prove that $B{D}^2-C{D}^2=2CD.AD$

Answer 1

Given $AB=AC,BD\perp AC$

From right $△ADB$,

By pythagorus theorem, $A{B}^2=A{D}^2+B{D}^2$

$⟹A{C}^2=A{D}^2+B{D}^2$ [$\because AB=AC$]

$⟹(AD+DC{)}^2=A{D}^2+B{D}^2$

$⟹A{D}^2+2.AD.DC+D{C}^2=A{D}^2+B{D}^2$

$⟹2.AD.DC=B{D}^2-C{D}^2$

Hence, proved