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Construction of a Quadrilateral in Different Situations

Let A B C be ...

Question

Let ABC be a triangle and P be an interior point. Prove that AB + BC + CA < 2(PA + PB + PC).

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Solution

Given: P is an interior point

To Prove: AB + BC + CA < 2(PA + PB + PC)

Proof:

In ΔAPB, AB < PA + PB … (1) (Sum of two sides of a triangle is greater than the third side)

In ΔCPB, BC < PC + PB … (2) (Sum of two sides of a triangle is greater than the third side)

In ΔAPC, AC < PA + PC … (3) (Sum of two sides of a triangle is greater than the third side)

Adding (1), (2) and (3):

AB + BC + CA < PA + PB + PC + PB + PA + PC

⇒ AB + BC + CA < 2(PA + PB + PC)

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