Two distinct points $A$ and $B$ lie on the same side of the plane divided by a line $L$. Let M be a point on L such that minimizes the value $MA + MB$. Let N be a point on L such that $NA = NB$. Show that $A$ , $B$ , $M$ , $N$ lies on a circle.
I tried reflect $A$ and $B$ over the line $L$ and draw a perpendicular line from $N$ to $AB$ so that it cuts AB in half with a right angle. Next , I tried angle chasing but it doesn’t work yet. So, I tried Menelaus’ Theorem and Ceva’s Theorem but again , it doesn’t work. Can anyone give me a hint or a solution of this problem? Thank you.
Say that point C is the reflection of point A with respect to line L. Obviously, point M is the intersection point of lines L and BC. Triangle AMC is isosceles because AM=MC. It follows that:
$$\angle MAC=\angle MCA=\angle BCA=\alpha, \angle AMB=2\alpha \tag{1}$$
It is also obvious that NA=NB=NC so points A,B,C are on a circle with radius NA and center N. Central angle ANB is twice the size of inscribed angle ACB:
$$\angle ANB=2\angle ACB=2\alpha\tag{2}$$
From (1) and (2):
$$\angle AMB=\angle ANB$$
So quadrangle AMNB is concyclic.