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The circle is touching the side BC of ABC at p and touching AB and AC produced at Q and R respectively proved that AQ 1/2 (perimeter of ABC)

Posted by Faizan Hussain 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

☆•..¤( Prateek )¤..•☆ 5 years, 1 month ago

Given: A 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 Hope it will help you ?✌

Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

Given: A 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... Hope it would be helpful to you ?✌

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