Given points W,X,Y,Z. Where the lengths, |WX| = |WZ| and |YX| = |YZ|. Let M be the midpoint of [X,Z]. Prove that the intersection of [W,Y] and [X,Z] is M.
So my approach for this started with me verifying that the angles $\angle WMX$, $\angle WMZ$, $\angle YMX$, $\angle YMZ$ are all the same, i.e. $90^\circ$. This was easy to prove, it just showing that triangles $\triangle WMX$ and $\triangle WMZ$ were congruent (Due to them sharing sides of the same lengths). Knowing that $\angle WMZ + \angle WMX = 180^\circ$, and that the two angles were the same, it is evident that each angle is $90^\circ$. Similar logic can be applied to angles $\angle YMX$, $\angle YMZ$.
What I dont know is how to show [W,Y] and [X,Z] intersect at M. Any ideas?
$\angle WMX=\angle YMX=90$ leads to $\angle WMY=180$.
Thus, W, M and Y are colinear, meaning M is on the line WY. Since M is also on the line XZ, they intersect at the point M.