Question

In an isoscles triangle ABC WITH AB=AC and BD is perpendicular to AC . Prove that BD square - CD square =2CD×AD

Answer 1

in ∆ABD,

AB^2=BD^2+DA^2

AC^2=BD^2+AD^2

(AD+DC)^2=BD^2+AD^2

AD^2+CD^2+2*AD*CD = BD^2+AD^2

CD^2+2 *AD* CD= BD^2

2CD*AD=BD^2 - CD^2

hence proved...

hope it helped...