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The incircle of triangle ABC touches …

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The incircle of triangle ABC touches the sides BC,CA,AB at D,E,F respectively. Show that AF+BD+CE=AE+BF+CD=1÷2 perimeter of triangle ABC

Posted by Last Bencher 6 years, 10 months ago

CBSE > Class 10 > Mathematics

1 answers

Garima Choudhary 6 years, 10 months ago

Perimeter of triangle ABC =AB+BC+AC, We know that AB= AF+BF , BC =BD+ CD, AC= AE+ CE, but the tangents are equal so, AF=AE_ _ _ _ _ (1) BD=BF_ _ _ _ _ (2) CE=CD_ _ _ _ _(3) Add eq. 1, 2 and 3 we get , AF+BD+CE= AE+BF+CD first condition proved . We know that perimeter of triangle ABC , AB+BC+CA=P AF+BF+BD+CD+AE+CE =P AF +AF+BD+BD+CE+CE =P 2(AF+BD+CE)= P SO, AF +BD+CE=P/2 i.e half of perimeter . Hence proved .

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