Question

Question #8a625

Answer 1

This can be proved as follows:

Explanation:

Consider any triangle ABC with a median BD drawn to the opposite side AC. It is required to be proved that AB+BC> BD

Consider triangles ABD and BCD. In these triangles AB> AD and BC> DC, because in a case otherwise, these triangles will not exist.

Now consider the triangle inequalities AB +AD>BD and BC + DC>BD. These inequalities would still hold good if AD id replaced by AB. This would make AB+AB> BD, that is 2AB>BD. Like wise 2BC >BD.

Now add both these inequalities 2AB +2BC > 2BD.

Therefore AB+BC > BD