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#### Question

In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = `\frac { 1 }{ 2 } ("perimeter of ∆ABC")`

#### Solution

#### Similar questions

Prove that the paralleogram circumscribing a circle, is a rhombus

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(A) 30°

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