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Question
In following figure , the incircle of Δ ABC , touches the sides BC , CA and AB at D , E and F respectively. Show AF + BD + CE = AE + BF + CD
Solution
To prove:- AF + BD + CE = AE + BF + CD
Proof:- AF = AE ----(1) {Length of tangents drawn from an external point to a circle are equal }
BD = BF ----(2}
CE = CD ----(3}
Adding (1), {2} and {3}
AF + BD + CE = AE + BF + CD
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