In an isosceles triangle ABC with AB = AC, BD is perpendicular from B to side AC. Prove that `BD^(2)-CD^(2)=2CD.AD`

In an isosceles triangle ABC with AB=ACBD is perpendicular from B to the side AC. Prove that BD^(2)-CD^(2)=2CDAD

In an isosceles triangle ABC with AB=AC,BD is perpendicular from B to the side AC. Prove BD^(2)-CD^(2)=2CD.AD

DeltaABC is an isosceles triangle. AB = AC and BD is perpendicular on AC from vertex B, then BD^(2)-CD^(2) is equal to-

In a right triangle ABC, right angled at A, AD is drawn perpendicular to BC. Prove that: AB^(2)-BD^(2)= AC^(2)-CD^(2)

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is the bisector of angleA . Prove that: BD = CD

ABC is an isosceles triangle with AB=AC and D is a pooint on AC such that BC^(2)=ACxCD .Prove that BD=BC

In Figure,ABC is an isosceles triangle with AB=AC,BD and CE are two medians of the triangle.Prove that BD=CE

In Delta ABC, AB= AC . Side BC is produced to D. Prove that (AD^(2)-AC^(2))=BD*CD .

In an isoscles Delta ABC, AB= AC and D is a point on BC. Prove that AB^(2)-AD^(2)=BD*CD .

In triangle ABC,/_BAC=90^(@) and AB=AC. Seg AP is perpendicular to side BC.D is any point on side BC.Prove that 2(AD)^(2)=(BD)^(2)+(CD)^(2)