If a circle is touching the side BC of △ ABC at P and is to

1.	If a circle is touching the side BC of △ ABC at P and is touching AB and AC produced at Q and R respectively (see the figure).Prove that AQ = \(\frac12\) (perimeter of △ ABC).
Answer» Given: A circle touching the side BC of △ABC at P and AB,AC produced at Q and R respectively. Proof: Lengths of tangents drawn from an external point to a circle are equal. ⇒ AQ = AR, BQ = BP, CP = CR. Perimeter of △ABC = AB + BC + CA = AB + (BP + PC) + (AR – CR) = (AB + BQ) + (PC) + (AQ – PC) since [AQ = AR, BQ = BP, CP = CR] = AQ + AQ = 2AQ ⇒ AQ = \(\frac12\)(Perimeter of △ABC) ∴ AQ is the half of the perimeter of △ABC

If a circle is touching the side BC of △ ABC at P and is touching AB and AC produced at Q and R respectively (see the figure).Prove that AQ = \(\frac12\) (perimeter of △ ABC).

Answer»

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

Proof: Lengths of tangents drawn from an external point to a circle are equal.

⇒ AQ = AR, BQ = BP, CP = CR.

Perimeter of △ABC = AB + BC + CA

= AB + (BP + PC) + (AR – CR)

= (AB + BQ) + (PC) + (AQ – PC) since [AQ = AR, BQ = BP, CP = CR]

= AQ + AQ

= 2AQ

⇒ AQ = \(\frac12\)(Perimeter of △ABC)

∴ AQ is the half of the perimeter of △ABC

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If a circle is touching the side BC of △ ABC at P and is touching AB and AC produced at Q and R respectively (see the figure).Prove that AQ = \(\frac12\)​ (perimeter of △ ABC).

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If a circle is touching the side BC of △ ABC at P and is touching AB and AC produced at Q and R respectively (see the figure).Prove that AQ = \(\frac12\) (perimeter of △ ABC).