Question

(a) A circle touches the sides AB, BC and CA of a triangle ABC at P, Q and R respectively
Show that
AP + BQ + CR = PB + QC + RA = 1/2(Perimeter of triangleABC).​

Answer 1

Step-by-step explanation:

refer to the attachment

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Answer 2

The image is attached below.

Given data:

AB, BC and CA are sides of a triangle.

A circle touches the sides AB, BC and CA at P, Q and R respectively.

Tangents drawn from an external point to a circle are equal.

AP and AR are tangents from the external point A.

AP = AR – – – – (1)

BQ and BP are tangents from the external point B.

BQ = BP – – – – (2)

CR and CQ are tangents from the external point A.

CR = CQ – – – – (3)

Adding (1), (2) and (3), we get

AP + BQ + CR = AR + BP + CQ – – – (4)

Perimeter of triangle ABC = AB + BC + CA

                                           = AP + BP + BQ + CQ + CR + AR

                                           = (AP + BQ + CR) + (AR + BP + CQ)

                                           =  (AP + BQ + CR) + (AP + BQ + CR) [From eqn (4)]

Perimeter of triangle ABC  = 2 (AP + BQ + CR)

$\frac{1}{2}$ (Perimeter of triangle ABC) = (AP + BQ + CR)

Similarly, $\frac{1}{2}$ (Perimeter of triangle ABC) = (AR + BP + CQ)

Hence AP + BQ + CR = PB + QC + RA = $\frac{1}{2}$ (Perimeter of triangle ABC).

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