In the square ABCD, M is the midpoint of AB.
A line is drawn through M perpendicular to CM to locate N.
Prove that the size of angle BCM is equal to the size of angle MCN.
We begin by drawing a line through M and parallel to BC to produce E on CN and F on CD.
As right angle triangle CMN is half of a rectangle, CN is one diagonal, and because E is the midpoint of CN, CE = EN = EM.
In which case, triangle CEM is isosceles, a = b (base angles equal).
Because MF is parallel to BC, a = c (alternate angles).
Hence b = c, and we prove that the size of angle BCM is equal to the size of angle MCN.
If AB=4, find the perimeter of triangle CDN.