In triangle ABC, given below, P is a point inside △ABC. Prove that : AP + BP + CP > (AB + BC + CA)

Given: P is a point inside △ ABC. To prove: AP + BP + CP > (AB + BC + CA) Proof: Using the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the third side: In Δ PAB: AP + BP > AB ...................(1) Similarly, in Δ PBC: BP + CP > BC ...................(2) Similarly, in Δ PAC: AP + CP > AC ...................(3) Now, adding the equations (1), (2) and (3): AP + BP + BP + CP + AP + CP > AB + BC + AC ⇒ 2AP + 2BP + 2CP > AB + BC + AC ⇒ 2(AP + BP + CP) > AB + BC + AC ⇒ AP + BP + CP > (AB + BC + AC) Hence, AP + BP + CP > (AB + BC + AC).

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Mathematics

In triangle ABC, given below, P is a point inside △ABC. Prove that : AP + BP + CP > (AB + BC + CA)

Triangles

Related Questions

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