Question

A circle is touching the side BC of ∆ABC at P and touching AB and AC at Q and R respectively prove that AQ = 1/2 ( Perimeter of ∆ABC )

Answer 1

Answer:

Answer Expert Verified

GIVEN: A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R. [Lengths of tangents drawn from an external point to a circle are equal.] Hence, AQ is half the perimeter of ΔABC.Jan 4, 2017

Answer 2

Step-by-step explanation:

circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively. RTP: AP = 1/2 (Perimeter of ΔABC) Proof: Lengths of tangents drawn from an external point to a circle are equal. ⇒ AQ = AR, BQ = BP, CP = CR. Perimeter of ΔABC = AB + BC + CA = AB + (BP + PC) + (AR – CR) = (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR] = AQ + AQ = 2AQ ⇒ AQ = 1/2 (Perimeter of ΔABC) ∴ AQ is the half of the perimeter of ΔABC.

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