Given below are two statements :
Statement I: The perimeter of a triangle is greater than the sum of its three medians.
Statement II: In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2 AD.
In the light of the above statements, choose the correct answer from the options given below :

Statement I: The perimeter of a triangle is greater than the sum of its three medians.

This is true. In a triangle:

The medians connect a vertex to the midpoint of the opposite side, dividing the triangle into two smaller triangles of equal area.

The length of each median is always less than the sum of the two sides it connects (triangle inequality).

Therefore, the sum of the three medians is always less than the perimeter of the triangle.

Statement II: In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD.

This is true as well. By the triangle inequality:

For the triangle formed by

AD, AB, and BD, we have

AB+BD>AD.

Similarly, for the triangle formed by

AD, AC, and CD, we have

AC+CD>AD.

Adding these inequalities gives

AB+AC+BD+CD>2AD.

Since

BD+CD=BC, we conclude

AB+AC+BC>2AD.

Statement I relies on the fact that medians are always shorter than the sum of the sides they span.

Statement II is a direct result of the triangle inequality applied to segments involving AD, proving its validity.

Therefore, the correct answer is option A.