Prove that $MN=\frac{1}{2}(AB+BC+CA)$

Question

As I was doing a question on quadrilateral, I found out this one, which I couldn't find to relate with any properties of a quadrilateral. It seems there is a trapezium in between, but it's quite unclear to me how to establish a relation. Please help if you can.

A diagram of the image

It is not given that:

1. $ABC$ is equilateral.

2. $MN$ bisects angle $M$ or angle $N$.

Answer 1

Let the lines AM and AN cut the line BC at P and Q, respectively Since triangle ABP(AB=BP) and triangle ACQ(AC=CQ) are isosceles triangles, M and N are the midpoint of AP and AQ, respectively. Then by Midpoint Theorem $$MN=\frac{1}{2}PQ=\frac{1}{2}(PB+BC+CQ)=\frac{1}{2}(AB+BC+CA)$$