1. Given: $ABCD$ is a square, and $M$ and $N$ are points on sides $AB$ and $BC$ respectively such that $\angle MDN = 45^\circ$. $R$ is the midpoint of $MN$. We need to show that $RP = RQ$ where $P$ and $Q$ are points of intersection of $AC$ with the lines $MD$ and $ND$ respectively.

2. Step 1: Since $ABCD$ is a square, all its sides are equal and all its angles are $90^\circ$. Let the side length of the square be $a$.

3. Step 2: Place the square in the coordinate plane with $A = (0, 0)$, $B = (a, 0)$, $C = (a, a)$, and $D = (0, a)$. Let $M = (x, 0)$ and $N = (a, y)$ for some $0 \leq x, y \leq a$.

4. Step 3: The coordinates of $D$ are $(0, a)$. The slope of $MD$ is $\frac{a - 0}{0 - x} = -\frac{a}{x}$. The slope of $ND$ is $\frac{a - y}{0 - a} = \frac{y - a}{a}$.

5. Step 4: The equation of line $MD$ is $y = -\frac{a}{x}x + a$. The equation of line $ND$ is $y = \frac{y - a}{a}x + a$.

6. Step 5: The diagonal $AC$ has the equation $y = x$ since it passes through $(0, 0)$ and $(a, a)$.

7. Step 6: To find the intersection points $P$ and $Q$, solve the equations of $MD$ and $ND$ with $AC$:
- For $P$: $x = -\frac{a}{x}x + a$ simplifies to $x = \frac{a}{2}$, so $P = \left(\frac{a}{2}, \frac{a}{2}\right)$.
- For $Q$: $x = \frac{y - a}{a}x + a$ simplifies to $x = \frac{a}{2}$, so $Q = \left(\frac{a}{2}, \frac{a}{2}\right)$.

8. Step 7: Since $P$ and $Q$ both lie on the line $AC$ and have the same coordinates, $P = Q$.

9. Step 8: The midpoint $R$ of $MN$ is $\left(\frac{x + a}{2}, \frac{y}{2}\right)$.

10. Step 9: Since $P = Q$, the distances $RP$ and $RQ$ are equal because they are both the distance from $R$ to the same point $P$.

Therefore, $RP = RQ$.

$\blacksquare$