Question

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

Answer 1

Answer:

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=

2

1

(BC)

Now, By Pythagoras theorem in ΔADX and ΔACX,

AD

2

=AX

2

+XD

2

and AC

2

=AX

2

+CX

2

⇒AD

2

−AC

2

=XD

2

−CX

2

=(BD−BX)

2

−CX

2

=BD

2

+BX

2

−2BD.BX−CX

2

(BX=CX)

=BD(BD−2BX)

=BD(BD−BC)

⇒AD

2

−AC

2

=BD⋅CD

∴AD

2

=AC

2

+BD⋅CD

solution

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