In the given figure. P is a point on the side BC of ∆ABC. Proce that (AB + BC + CA) > 2AP.

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Solution

Sum of any two sides of a triangle is greater than the third side.

$I n △ A P B : A B + B P > A P I n △ A P C : A C + P C > A P A d d i n g t h e c o r r e s p o n d o n g s i d e s : A B + B P + A C + P C > A P + A P A B + A C + B C > 2 A P$

Hence, proved.

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