Math, asked by Anonymous, 1 year ago

Solve for it....... ​

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Answered by sarohakaran79
0

Answer:

Step-by-step explanation:

Sol:

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

to prove: AQ = 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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Answered by saswat2084
0

Here is the proof of ur question buddy

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