Each side of square ABCD has length 1 unit. Points P and Q are on AB, and DA, respectively. Find angle PCQ if the perimeter of triangle APQ is 2.
I used trigonometry to solve this problem, and my answer is 45 degrees. I would like to know if there are any other, more elegant methods to solve this problem.
Notice that since the perimeter of $APQ$ is $2$ (i.e., half of that of the square), we have $AB+AD = AQ+AP+PQ$ and $AB+AD = AQ+QD +AP+PB$. Combining these two equations obtain, $PQ = QD+PB$. Thus, there is a point $X$ on $PQ$ such that $PX = PB$ and $QX = QD$. With this in mind, extend $PB$ to a point $W$ in the same direction but with $PW = PQ$. Similarly extend $QD$ to a point $V$ such that $QV = PQ$. Now let angle $QCD = \theta$ and angle $PCB = \alpha$. Notice that triangle $PQC$ is congruent to triangles $QCV$ and $PCW$ and also that triangle $DCV$ is congruent to triangle $PBC$. Hence it follows that the angle $PCQ$ is $\theta +\alpha$, and that $2\theta+2\alpha = 90^\circ$ which gives the required $45$ degree angle.