Question

ΔABC is an isosceles triangle in which AB ≅ AC. A circle touching all the three sides of ΔABC touches BC at D. Prove that D is the mid-point of BC.

Answer 1

drawn an isosceles triangle ABC in which AB ≌ AC also a circle touching all three sides of ∆ABC touches BC at D, CA at E and AB at F.

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

proof :- $\textbf{As we know that if a circle touches all sides}\\\textbf{of a triangle, the lengths of tangents}\\\textbf{ drawn from each vertex are equal}$.
∴ AE = AF............(i)
BD = BF ...............(ii)
and CD = CE …..........(iii)
Consider AB = AC,
Subtracting AF from both sides,
⇒ AB – AF = AC – AF
From equation. (i),
⇒ AB – AF = AC – AE
Since A – F – B and A – E – C,
⇒ BF = CE
From equations. (ii) and (iii),
⇒ BD = CD
hence, D is the midpoint of BC.