Question

AM is the median on side BC of AABC. Prove that A(AABM) = A(AACM).​

Answer 1

Answer:

We know,

In a triangle sum of two sides is greater than the third side

Hence,In triangle ABM

AB + BM > AM 1

Also in triangle ACM

AC + MC > AM 2

Adding equation 1 and 2 we get,

AB + BM + AC + MC > AM + AM

AB + AC + (BM + MC) > 2AM

AB + AC + BC > 2AM