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A circle is touching the side …

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A circle is touching the side BC of a triangle ABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ is half the perimeter of triangle ABC
  • 1 answers

Sia ? 6 years, 6 months ago


We know that the two tangents drawn to a circle from an external point are equal.
{tex}\therefore{/tex} AQ = AR, BP = BQ, CP = CR
{tex}\therefore{/tex} Perimeter of {tex}\triangle{/tex}ABC =  AB + BC + AC
= AB + BP + PC + AC
= AB + BQ + CR + AC [{tex}\because{/tex} BP = BQ, PC = CQ]
= AQ + AR = 2AQ = 2AR = [{tex}\because{/tex} AQ = AR]
= AQ = AR = {tex}\frac 12{/tex}[Perimeter of {tex}\triangle{/tex}ABC]

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