Question

In $△ABC$, $AB=AC$. Side $BC$ is produced to $D$. Prove that ${AD}^2-{AC}^2=BD\times CD$

Answer 1

Applying Pythagoras theorem in right-angled $△AED$, we get:

Since, $ABC$ is an isosceles triangle and $AE$ is the altitude and we know that the altitude is also the median of the isosceles triangle.

So, $BE=CE$

And $DE+CE=DE+BE=BD$

${AD}^2={AE}^2+{DE}^2$

$\Rightarrow {AE}^2={AD}^2-{DE}^2$ ........$\left(1\right)$

In $△ACE$,

${AC}^2={AE}^2+{EC}^2$

$\Rightarrow {AE}^2={AC}^2-{EC}^2$ ........$\left(2\right)$

Using $\left(1\right)$ and $\left(2\right)$

$\Rightarrow {AD}^2-{AC}^2={DE}^2-{EC}^2$

$=(DE+EC)(DE-EC)$

$=(DE+BE)\times CD$

$=BD\times CD$

$\therefore {AD}^2-{AC}^2=BD\times CD$

Hence proved.