The incircle of an isosceles triangle $A B C$ , in which $A B = A C$ , touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$ respectively. Prove that $B D = D C .$

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Solution

Given: $Δ A B C$ is isosceles triangle in which $A B = A C$ , with a circle inscribed in a triangle.

To prove: $B D = D C$

Proof: $A F$ and $A E$ are tangents drawn to the circle from point $A$ .

Since two tangents drawn to a circle from the same exterior point are equal, $A F = A E = a$

Similarly, $B F = B D = b$ , and $C D = C E = c$

$Δ A B C$ is a isosceles triangle in which $A B = A C$

So, $a + b = a + c$

Thus, $b = c$

Therefore, $B D = D C$

Hence proved.

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The incircle of an isosceles triangle ABC, in which AB=AC, touches the sides BC,CA and AB at D,E and F respectively. Prove that BD=DC.

The incircle of an isosceles triangle $A B C$ , in which $A B = A C$ , touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$ respectively. Prove that $B D = D C .$