Prove that the sum of the distances of any point within a triangle , from its vertices , is less than the perimeter of the triangle .

Abhinav Sharma Grade: 11

        Prove that the sum of the distances of any point within a triangle , from its vertices , is less than the perimeter of the triangle .

7 years ago

Answers : (2)

Aman Bansal

592 Points

Dear Abhinav Sharma,

consider an equlateral triangle abc,

Let us draw lines DE parallel to AC, FG parallel to AB, IJ parallel to BC. Triangles FPJ, DIP, and PGE are similar to ABC and therefore are equilateral.

Denote the heights of the triangles FPJ, DIP, PGE, and ABC by e, f, g, h, respectively. e/h = FJ/AC, f/h = DI/AB, g/h = GE/BC. Let us add these thee equalities. Note that DP = AF and PE = JC because ADPF and JPEC are parallelograms. Using the fact that three sides of each of FPJ, DIP, PGE, and ABC are equal, we get the following equality: (e+f+g))/h = 1.

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AMAN BANSAL

7 years ago

Abhinav Sharma

29 Points

But you didnt answer my question , I asked to prove the following :

Prove that : AP + PC + PB < AB + BC + AC . ( Point lying anywhere within a triangle ).