In the figure, show that perimeter of $△ A B C = 2 (A P + B Q + C R) .$

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Solution

Perimeter of $△ A B C = A B + B C + A C$
$= A P + P B + B Q + Q C + C R + R A$
but AP = AR (tangents drawn from A to the circle)
PB = BQ (tangents drawn from B to the circle)
CQ = CR (tangents drawn from C to the circle)
$∴$ Perimeter of $△ A B C = A P + B Q + B Q + C R + C R + A P$
$= 2 A P + 2 B Q + 2 C R$
$= 2 (A P + B Q + C R)$

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Please help me with the following

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

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In the figure, show that perimeter of △ABC=2(AP+BQ+CR).

Perimeter of △ABC=AB+BC+AC=AP+PB+BQ+QC+CR+RAbut AP = AR (tangents drawn from A to the circle) PB = BQ (tangents drawn from B to the circle) CQ = CR (tangents drawn from C to the circle)∴ Perimeter of △ABC=AP+BQ+BQ+CR+CR+AP =2AP+2BQ+2CR =2(AP+BQ+CR)

In the figure, show that perimeter of $△ A B C = 2 (A P + B Q + C R) .$