Prove that the sum of the distances of any point within a triangle , from its vertices, is smaller than the perimeter of the triangle .(As shown in the following figure )Prove that : AP + PC + PB < AB + BC + AC . ( Point P lying anywhere within the triangle ).

Dear student,

Lets take a different view of this problem muh more interesting and relevant as far as JEE is concerned...

Fermat point: Another center is the Fermat point (also called the isogonic center or the Rorricelli point). This is the point which minimizes the sum of the distances from the three vertices. This problem is called Fermat's problem or Steiner's problem. The three angles at this point, between the vertices, are all 120 degrees. This point has many practical uses, as it is often a good idea to minimize distances, so save money. Telephone lines between three cities, using the minimum amount of wire, will be three lines which meet at the Fermat point. More points result in connections between Fermat points. The Fermat point can be constructed by constructing equilateral triangles on each side of the triangle, and connecting their farthest vertices with the opposite vertex of the original triangle.

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