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Question
In following fig., a circle is touching the side BC of Δ ABC at P and AB and AC produced at Q and R respectively. Prove that AQ is half the perimeter of Δ ABC.
Solution
To prove:· AQ = `1/2` (Perimeter of Δ ABC)
Proof : BQ = BR = 5 - r ....(1) PC = CR = 12 - r .....(2) ] (1)
(Lengths of tangents drawn from an external point to a circle are equal)
Perimeter of 6 ABC = AB + BC+ AC
= AB + BP + PC + AC
= AB+BQ+CR+AC Using (1)
= AQ+AR
= 2 AQ
2 AQ = Perimeter of Δ ABC
AQ = `1/2` (Perimeter of Δ ABC)
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