InterviewSolution
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InterviewSolution
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In the given figure, D, E and F are the points where the incircle of the `Delta`ABC touches F the sides BC, CA and AB respectively. Show that AF+ BD + CE = AE + BF + CD = (perimeter of `Delta ABC`) |
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Answer» We know, length of tangents from any common point outside a circle are always equal. `:. AF = AE ->(1)` `BD = BF->(2)` `CE = CD ->(3)` Adding (1),(2) and (3), `AF+BD+CE = AE+BF+CD` Now, perimeter of `ABC = AB+BC+AC` `=AF+BF+BD+DC+CE+AE` `=2(AF+BD+CE)` `:. 1/2 (` Perimeter of `ABC) = AF+BD+CE = AE+BF+CD.` |
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