Find the measure of angle $MAN$. We construct two points $M,N$ such that perimeter of triangle $MCN$ is equal to half of the perimeter of the square.Find the measure of angle $MAN$.
The case $CM=CN$ leads to a simple equation so we can find the measure of $MAN$ using trigonometry.But in general case I don't no How to use equality of perimeter maybe we have to construct the same triangle in another place and use the sum of their perimeter.But I didn't get anything from this method.Any hints?

I think I understood about which square you say! 
I think your problem is the following.

Let $ABCD$ is a square with $AB=a$. $M\in BC$ and $N\in CD$ such that $CN+CM+MN=2a$. Find $\measuredangle MAN$.

Let $MC=x$ and $CN=y$.
Hence, $$x+y+\sqrt{x^2+y^2}=2a,$$
$$AM=\sqrt{(a-x)^2+a^2}=\sqrt{2a^2-2ax+x^2}=\sqrt{\frac{1}{2}\left(x+y+\sqrt{x^2+y^2}\right)^2-\left(x+y+\sqrt{x^2+y^2}\right)x+x^2}=$$
$$=\sqrt{\sqrt{x^2+y^2}\left(y+\sqrt{x^2+y^2}\right)}.$$
By the same way we obtain:
$$AN=\sqrt{\sqrt{x^2+y^2}\left(x+\sqrt{x^2+y^2}\right)}.$$
Thus,
$$\measuredangle MAN=\arccos\frac{AM^2+AN^2-MN^2}{2AM\cdot AN}=$$
$$=\arccos\frac{4a^2-2a(x+y)+x^2+y^2-x^2-y^2}{2\sqrt{x^2+y^2}\sqrt{\left(y+\sqrt{x^2+y^2}\right)\left(x+\sqrt{x^2+y^2}\right)}}=$$
$$=\arccos\frac{2a(2a-x-y)}{2\sqrt{x^2+y^2}\sqrt{x^2+xy+y^2+(x+y)\sqrt{x^2+y^2}}}=$$
$$=\arccos\frac{a}{\sqrt{2a^2}}=\arccos\frac{1}{\sqrt2}=45^{\circ}.$$
Done!
