Finding angle with pure Geometry. 
Each side of a square $ABCD$ has a length of $1$ unit. Points $P$ and $Q$ belong to $AB$ and $DA$ respectively. The perimeter of triangle $APQ$ is $2$ units. What will be the angle of $PCQ$?

I was able to do this with simple trigonometry and found it to be $45^\circ$ but the book I am reading doesn't yet talk about trigonometry or similarity of triangles so I want rather pure geometric proof which doesn't include trigonometry or similarity of triangle concepts.
 A: (I apologize for replacing points $PQ$ with $EF$, I'm too tired to draw the picture again :)
Draw circle with center $E$ and radius $EB$ and circle with center $F$ and radius $FD$. These circles intersect segment $EF$ at points $G',G''$ These points are identical! Why? 
Because perimeter of triangle $AEF$ is 2 which is  $AB+AD=AE+EB+AF+FD=AE+EG'+AF+FG''$. It means that $EG'+FG''=EF$ and therefore $G'\equiv G''\equiv G\in EF$. 
So these two circles touch at point $G$ as shown in the picture. Power of point $C$ with respect to both circles is equal $(CB=CD)$ and therefore it has to be on the radical axis of the circles. These circles touch at point $G$ and their radical axis is defined by the common tangent at point $G$. Becuase of that $CG$ must be tangent to both circles and, at the same time, $CG\bot EF$.
The rest is trivial: you can easily show that triangles $FCD$ and $FCG$ are congruent. The same is true for triangles $ECG$ and $ECB$. Because of that:
$$\angle ECF=\frac12\angle BCG+\frac 12\angle GCD=45^\circ$$

A: Rotate $Q$ for $90^{\circ}$ around $C$ in to new point $E$. Then $P,B,E$ are collinear and $PE = PQ$. So triangles $CQP$ and $CEP$ are congruent by (sss), so $$\angle QCP = \angle ECP = 45^{\circ}$$
