# ΔABC is an isosce

Given that ΔABC is an isosceles triangle in which AB AC and a circle touching all the three sides of ΔABC touches BC at D.

We have to prove that D is the midpoint of BC.

Proof:

We know that if a circle touches all sides of a triangle, the lengths of tangents drawn from each vertex are equal.

AE = AF, BD = BF and CD = CE … (1)

Consider AB = AC,

Subtracting AF from both sides,

AB – AF = AC – AF

From (1),

AB – AF = AC – AE

Since A – F – B and A – E – C,

BF = CE

From (1),

BD = CD

Since B – D – C and BD = CD,

D is the midpoint of BC.

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