In the figure below, the lines AB, BC, CA touch the circle at P, Q, R.

Prove that the perimeter of the triangle is 2(AP + BQ + CR)

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Solution

We know that perimeter of a triangle is equal to the sum of the lengths of its sides.

Perimeter of ΔABC = AB + BC + CA

= (AP + PB) + (BQ + QC) + (AR + RC) …(1)

We know that lengths of tangents drawn from a point outside a circle are equal.

⇒ AP = AR ... (2)

BP = BQ ... (3)

CR = CQ ... (4)

Substituting the values from equations (2), (3) and (4) in equation (1):

Perimeter of ΔABC = AP + BQ + BQ + CR + AP + CR

= 2(AP + BQ + CR)

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The sides of a triangle touch a circle. Sides AB, BC, CA touch the circle at points P, Q, R respectively. Prove that:

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

2.AP +BQ+CR=1/2 *perimeter of 🔺 ABC

In the given figure, the incircle of $△ A B C$ touches the sides AB, BC and CA at the points P, Q, R respectively. Show that
$A P + B Q + C R = B P + C Q + A R$
$= \frac{1}{2} (P e r i m e t e r o f △ A B C)$
[3 MARKS]

Q. In the figure, show that perimeter of

△ A B C = 2 (A P + B Q + C R) .

Q. If the incircle of

△ A B C

touches the sides

A B, B C

and

C A

P, Q

and

R

, find the value of :

A P + B Q + C R

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In the figure below, the lines AB, BC, CA touch the circle at P, Q, R. Prove that the perimeter of the triangle is 2(AP + BQ + CR)

In the figure below, the lines AB, BC, CA touch the circle at P, Q, R.

Prove that the perimeter of the triangle is 2(AP + BQ + CR)