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,$BX=CX=\frac{2}{1}(BC)$

Now, By Pythagoras theorem in$ΔADX\:and\:ΔACX,$

$AD²=AX²+XD²$

and $AC²=AX² +CX²$

$⇒AD²−AC²=XD² −CX²$

$\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:=(BD−BX)² −CX²$

$\:\:\:\:\:\:\:\:\:\:\:\: \:\:\:\:\:\:\:\:\:\:\:\:=BD²+BX²−2BD.BX−CX²(BX=CX)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: =BD(BD−2BX)$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=BD(BD−BC)$

$⇒AD²−AC²=BD⋅CD$

$∴AD²=AC²+BD⋅CD$

Hence, proved.

Answer 2

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