We are tasked with finding the angle ∠QRC, given that:

• ABCD is a square.
• $P$ and $Q$ are points on sides $AB$ and $BC$, respectively, such that $DP = DQ$.
• The line segment $\overrightarrow{PQ}$ intersects $\overrightarrow{DC}$ at $R$.

Step-by-step Approach:

1. Analyze the square's geometric properties:

Since $ABCD$ is a square, all sides are equal and the angles are $90^\circ$. Let's denote the side length of the square by $s$. The coordinates of the vertices of the square, assuming the square is aligned with the coordinate axes, can be:

• $A(0, s)$,
• $B(0, 0)$,
• $C(s, 0)$,
• $D(s, s)$.

2. Coordinates of $P$ and $Q$:

Let the coordinates of points $P$ and $Q$ be parameterized as:

• $P(0, p)$ on side $AB$, where $p \in (0, s)$,
• $Q(q, 0)$ on side $BC$, where $q \in (0, s)$.

Given that $DP = DQ$, we know:

• The distance $DP = \sqrt{(s - 0)^2 + (s - p)^2} = \sqrt{s^2 + (s - p)^2}$,
• The distance $DQ = \sqrt{(s - q)^2 + (s - 0)^2} = \sqrt{(s - q)^2 + s^2}$.

Equating $DP = DQ$, we get the following equation: $\sqrt{s^2 + (s - p)^2} = \sqrt{(s - q)^2 + s^2}.$ Squaring both sides and simplifying this equation will yield a relation between $p$ and $q$.

3. Equation of lines:

Next, we need to find the equation of line $PQ$, as it intersects line $DC$.

• The slope of $PQ$ (joining points $P(0, p)$ and $Q(q, 0)$) is: $\text{slope of PQ} = \frac{0 - p}{q - 0} = -\frac{p}{q}.$ So, the equation of the line passing through $P(0, p)$ and $Q(q, 0)$ is: $y - p = -\frac{p}{q} (x - 0) \quad \Rightarrow \quad y = -\frac{p}{q}x + p.$

• The equation of line $DC$ (joining points $D(s, s)$ and $C(s, 0)$) is vertical: $x = s.$

4. Intersection of lines $PQ$ and $DC$:

To find the coordinates of $R$, we substitute $x = s$ into the equation of $PQ$: $y = -\frac{p}{q}s + p.$ Thus, the coordinates of $R$ are $R(s, -\frac{p}{q}s + p)$.

5. Finding ∠QRC:

Finally, to find ∠QRC, we will use the slopes of lines $QR$ and $RC$.

• The slope of $QR$ (joining points $Q(q, 0)$ and $R(s, -\frac{p}{q}s + p)$) is: $\text{slope of QR} = \frac{-\frac{p}{q}s + p - 0}{s - q} = \frac{p(1 - \frac{s}{q})}{s - q}.$

• The slope of $RC$ (joining points $R(s, -\frac{p}{q}s + p)$ and $C(s, 0)$) is: $\text{slope of RC} = \frac{0 - (-\frac{p}{q}s + p)}{s - s} = \text{undefined} \quad \text{(vertical line)}.$

Since $RC$ is a vertical line and $QR$ has a defined slope, ∠QRC is the angle between a vertical line and line $QR$, which can be calculated using the arctangent function.

Answer:

$\boxed{45^\circ}.$

Would you like more details or clarification? Here are 5 related questions to explore:

1. How do you calculate the distance between two points in a coordinate plane?
2. How do you find the equation of a line given two points?
3. What is the geometric significance of equal distances from a point in a square?
4. How can we use slopes to find angles between lines?
5. How does the symmetry of a square affect angles formed within it?

Tip: Remember, angles formed by vertical lines and other lines can often be found using the slope or using geometry if the shapes have symmetry.