From the given figure, prove that :

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

Also, show that :

AP + BQ + CR = $\dfrac{1}{2}$ x Perimeter of triangle ABC.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From point B, BQ and BP are the tangents to the circle

BQ = BP …….. (1)

From point A, AP and AR are the tangents to the circle

AP = AR …….. (2)

From point C, CR and CQ are the tangents to the circle

CR = CQ …….. (3)

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

AP + BQ + CR = BP + CQ + AR ……… (4)

Hence, proved that AP + BQ + CR = BP + CQ + AR.

Now, adding AP + BQ + CR to both sides in (4), we get

2(AP + BQ + CR) = AP + BP + CQ + BQ + AR + CR ……….(5)

From figure,

AP + BP = AB, BQ + CQ = BC and AR + CR = AC.

Substituting above value in equation (5), we get :

⇒ 2(AP + BQ + CR) = AB + BC + CA

⇒ AP + BQ + CR = $\dfrac{1}{2}$(AB + BC + CA).

⇒ AP + BQ + CR = $\dfrac{1}{2}$ (Perimeter of △ABC). [∵ Perimeter = AB + BC + CA]

Hence, proved that AP + BQ + CR = $\dfrac{1}{2}$ (Perimeter of △ABC).