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.
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
⇒ AB – AF = AC – AE
Since A – F – B and A – E – C,
⇒ BF = CE
⇒ BD = CD
Since B – D – C and BD = CD,
D is the midpoint of BC.
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