Question

In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD2 = AC2 + BD x CD.​

Answer 1

Let AX be a line perpendicular to BC and as perpendicular from a vertex common to equal side is also a median therefore, BX=CX= 1/2 (BC)

Now by Phythagoras theorem Δ ADX and Δ ACX

AD² = AX²+XD²

and AC² = AX² + CX²

= AD² - AC² = XD² - CX²

                  = (BD-BX)²- CX²

                 =BD²+BX²-2BD.BX-CX²       (BX=CX)

                  = BD (BD-2BD)

                  =BD(BD-BC)

ANSWER= AD²-AC²= BD.CD

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