In triangle ABC on side BC, point D is taken, which is connected to vertex A.

In triangle ABC on side BC, point D is taken, which is connected to vertex A. Prove that the perimeter of triangle ABC is greater than the perimeter of triangle ADC.

Let’s define the perimeters of triangles ABC and ADС.

Ravs = AC + AB + BC = AC + AB + ВD + СD = (AC + СD) + (AB + ВD).

Rads = (AB + СD) + AD.

At the perimeters, the terms (AC + СD) are the same, then it remains to prove that (AB + ВD)> AD.

By the condition of constructing triangles, the sum of the lengths of any two other sides is greater than the length of the third side.

Then in the triangle AВD (AB + ВD)> AD, which means Ravs> Rads, which was required to prove.