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A circle touches the side BC …

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A circle touches the side BC of a triangle ABC at P andAB and AC when produced at Q and R respectively, show that AQ=1/2( perimeter of triangle ABC

Posted by Aryan Dixena 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Given: A circle touches the side BC of a ΔABC at P and AB and AC produced at Q and R respectively.

To Prove: AQ = {tex}\frac 12{/tex} perimeter of {tex}\triangle{/tex}ABC
Proof: {tex}\because{/tex}Length of the two tangents fron an external point to a circle are equal
{tex}\therefore{/tex} AQ = AR ........(1)
BQ = BP ........ (2)
CP = CR .........(3)
{tex}\therefore{/tex} Perimeter of {tex}\triangle{/tex}ABC = AB + BC + AC
= AB + (BP + CP) + AC
= (AB + BQ)+ (CR + AC) [Using (2) and (3)]
= AQ + AR
= AQ - AR
= AQ + AQ [from (1)]
= 2AQ
{tex}\Rightarrow{/tex} AQ = {tex}\frac 12{/tex} perimeter of {tex}\triangle{/tex}ABC

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