Q. 25.0( 2 Votes )

ΔABC is an isosce

Answer :

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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