Question

In $\Delta$ ABC, AB = and BD is perpendicular to AC. Prove that :
$B{D}^2-C{D}^2=2CD\times AD$

Answer 1

In triangle ABC, If $AB=AC$, then $AB=AC=AD+CD$............(eqn 1)

Since BD is perpendicular to AC therefore, ABD is right angled.

In triangle ABD,

$A{B}^2={AD}^2+{BD}^2$

${\left(AD+CD\right)}^2={AD}^2+{BD}^2$ ( from equation 1)

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

$2AD\times CD+C{D}^2={BD}^2$

$2AD\times CD={BD}^2-C{D}^2$

Hence Proved.