Question

In a $\Delta ABC,AB=AC$. If $BD\perp AC$, prove that $B{D}^2-C{D}^2=2CD\times AD$.

Answer 1

In right angle triangle ABD,

$A{B}^2=B{D}^2+A{D}^2$(Pythagoras theorem)

but $AB=AC$

$A{C}^2=B{D}^2+A{D}^2$

$\Rightarrow (AD+DC{)}^2=B{D}^2+A{D}^2$

$\Rightarrow A{D}^2+D{C}^2+2AD\cdot DC=B{D}^2+A{D}^2$

$\therefore 2AD\cdot DC=B{D}^2-D{C}^2$.