$R$ is the midpoint of $MN$ and the points where $AC$ intersects $MD$ and $ND$ are $P$ and $Q$, respectively. Show that $PR=QR$. 
In square $ABCD$, $M$ and $N$ are points on $AB$ and $BC$, respectively such that $\angle MDN=45°$. $R$ is the midpoint of $MN$ and the points where $AC$ intersects $MD$ and $ND$ are $P$ and $Q$, respectively. Show that $PR=QR$. 

I'm actually trying to prove that $PQMN$ is cyclic and the circumcircles of $PQMN$ and $\Delta BMN$ are the same. Am I going on the right track? Please help.
 A: Yes, it is the right track. You may also directly prove that $PQNBM$ is a cyclic pentagon.
Assuming $D=(0,0),C=(1,0),N=(1,x_0)$, the equation of the $MD$-line is given by:
$$ y = \frac{1+x_0}{1-x_0}\,x $$
by the tangent addition formula, hence $M=\left(\frac{1-x_0}{1+x_0},1\right)$ and $R=\left(\frac{1}{1+x_0},\frac{1+x_0}{2}\right)$.
The equation of the $AC$-line is $y=1-x$, hence $P=\left(\frac{1}{1+x_0},\frac{x_0}{1+x_0}\right)$ and $Q=\left(\frac{1-x_0}{2},\frac{1+x_0}{2}\right)$.
Now it is straightforward to check that $RP=RQ=RN=RB=RM=\frac{1+x_0^2}{2(1+x_0)}$.

You may also prove that $\widehat{NPD}=\widehat{MQD}=\frac{\pi}{2}$, and everything follows nicely.
A: just to show things behind it.

A: My idea is more or less the same as @TomK 's.

First of all, we concentrate on the green circle (that will go through $A, M, Q, D$) . It is cyclic because $\alpha = \alpha_1 = \alpha_2 = 45^0$
Therefore, $\theta_1 + \phi_1 =  \theta_2 + \phi_2 = 90^0$
Together with R is the midpoint of MN, we have “$QMN$ is a semi-circle with $R$ as center”.
By going through the same process using the blue circle, you can arrive at the conclusion “$PMN$ is a semi-circle with $R$ as center”.
Result follows.
A: You can use the fact that angle QAM = QDM (=45) to show that AMQD is a circle. Hence DMQ is also 45.
Similarly DCNP is also a circle. Keep angle chasing and you should be able to conclude that MPQN is a circle and that the presence of a right angle at MQN reveals MN as the diameter as required.
