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Question
A circle touches the side BC of a ΔABC at a point P and touches AB and AC when produced at Q and R respectively. As shown in the figure that AQ = `1/2` (Perimeter of ΔABC).
Solution
We have to prove that
AQ = `1/2` (perimeter of ΔABC)
Perimeter of ΔABC = AB + BC + CA
= AB + BP + PC + CA
= AB + BQ + CR + CA
(∵ Length of tangents from an external point to a circle are equal ∴ BP = BQ and PC = CR)
= AQ + AR ...(∵ AB + BQ = AQ and CR + CA = AR)
= AQ + AQ ...(∵ Length of tangents from an external point are equal)
= 2AQ
`\implies` AQ = `1/2` (Perimeter of ΔABC)
Hence proved.
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