Q4 of 62 Page 193

A circle touches the side BC of ΔABC at P, AB and AC produced at Q and R respectively, prove that AQ = AR =( perimeter of ΔABC)


Lengths of tangents drawn from an exterior point to the circle are equal.


From the above theorem,


AR = AQ …(1)


BP = BQ …(2)


CP = CR …(3)


Now,


Perimeter of ΔABC = AB + BC + CA


⇒ Perimeter of ΔABC = AB + BP + PC + CA


⇒ Perimeter of ΔABC = (AB + BQ) + (CR + CA) [Using (2) and (3)]


⇒ Perimeter of ΔABC = AQ + AR


⇒ Perimeter of ΔABC = AQ + AQ


⇒ Perimeter of ΔABC = 2AQ [Using (1)]


∴ AQ = AR = 1/2Perimeter of ΔABC


Hence, Proved.


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