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प्रश्न
If a circle is touching the side BC of ΔABC at P and is touching AB and AC produced at Q and R respectively (see the figure). Prove that AQ = `1/2` (perimeter of ΔABC).
योग
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उत्तर
Given, circle touching the side BC of ΔABC at point P and AB, AC produced at Q and R, respectively.
We know, lengths of tangents drawn from on external point to a circle are equal.
∴ AQ = AR ...(i)
BQ = BP ...(ii)
CP = CR ...(iii)
Perimeter of ΔABC = AB + BC + CA
=AB + (BP + PC) + (AR – CR)
= (AB + BP) + PC + (AQ – CP) ...[From equations (i) and (ii)]
= (AB + BQ) + PC + (AQ – CP) ...[From equation (ii)]
= AQ + PC + AQ – PC
= 2AQ
∴ AQ = `1/2` (perimeter of ΔABC)
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