Problem 1699

1. Identify the coordinates of the vertices of the square:
Let the square $ABCD$ be positioned in the coordinate plane with vertices at:
$$A = (0, 0), \quad B = (a, 0), \quad C = (a, a), \quad D = (0, a)$$
where $a$ is the side length of the square.

2. **Determine the coordinates of points $E$ and $F$:**
Since $E$ is the midpoint of $BC$ and $F$ is the midpoint of $CD$, their coordinates are:
$$E = \left( a, \frac{a}{2} \right), \quad F = \left( \frac{a}{2}, a \right)$$

3. **Find the equations of lines $AF$ and $AE$:**
- The slope of $AF$ is:
$$\text{slope of } AF = \frac{a - 0}{\frac{a}{2} - 0} = 2$$
Thus, the equation of line $AF$ is:
$$y = 2x$$
- The slope of $AE$ is:
$$\text{slope of } AE = \frac{\frac{a}{2} - 0}{a - 0} = \frac{1}{2}$$
Thus, the equation of line $AE$ is:
$$y = \frac{1}{2}x$$

4. **Find the equation of diagonal $BD$:**
- The slope of $BD$ is:
$$\text{slope of } BD = \frac{a - 0}{0 - a} = -1$$
Thus, the equation of line $BD$ is:
$$y = -x + a$$

5. **Find the intersection points $P$ and $Q$:**
- Intersection point $P = AF \cap BD$:
$$2x = -x + a \implies 3x = a \implies x = \frac{a}{3}$$
Substituting $x = \frac{a}{3}$ into $y = 2x$:
$$y = 2 \left( \frac{a}{3} \right) = \frac{2a}{3}$$
Thus, $P$ has coordinates:
$$P = \left( \frac{a}{3}, \frac{2a}{3} \right)$$
- Intersection point $Q = AE \cap BD$:
$$\frac{1}{2}x = -x + a \implies \frac{3}{2}x = a \implies x = \frac{2a}{3}$$
Substituting $x = \frac{2a}{3}$ into $y = \frac{1}{2}x$:
$$y = \frac{1}{2} \left( \frac{2a}{3} \right) = \frac{a}{3}$$
Thus, $Q$ has coordinates:
$$Q = \left( \frac{2a}{3}, \frac{a}{3} \right)$$

6. **Verify that $P$ and $Q$ divide $BD$ into three equal segments:**
- The coordinates of $B$ and $D$ are $B = (a, 0)$ and $D = (0, a)$.
- The length of diagonal $BD$ is:
$$BD = \sqrt{(a - 0)^2 + (0 - a)^2} = \sqrt{a^2 + a^2} = a\sqrt{2}$$
- The distances $BP$, $PQ$, and $QD$ are:
$$BP = \sqrt{\left( a - \frac{a}{3} \right)^2 + \left( 0 - \frac{2a}{3} \right)^2} = \sqrt{\left( \frac{2a}{3} \right)^2 + \left( -\frac{2a}{3} \right)^2} = \sqrt{\frac{4a^2}{9} + \frac{4a^2}{9}} = \sqrt{\frac{8a^2}{9}} = \frac{2a\sqrt{2}}{3}$$
$$PQ = \sqrt{\left( \frac{2a}{3} - \frac{a}{3} \right)^2 + \left( \frac{a}{3} - \frac{2a}{3} \right)^2} = \sqrt{\left( \frac{a}{3} \right)^2 + \left( -\frac{a}{3} \right)^2} = \sqrt{\frac{a^2}{9} + \frac{a^2}{9}} = \sqrt{\frac{2a^2}{9}} = \frac{a\sqrt{2}}{3}$$
$$QD = \sqrt{\left( \frac{2a}{3} - 0 \right)^2 + \left( \frac{a}{3} - a \right)^2} = \sqrt{\left( \frac{2a}{3} \right)^2 + \left( -\frac{2a}{3} \right)^2} = \sqrt{\frac{4a^2}{9} + \frac{4a^2}{9}} = \sqrt{\frac{8a^2}{9}} = \frac{2a\sqrt{2}}{3}$$
- Since $BP = PQ = QD = \frac{a\sqrt{2}}{3}$, the diagonal $BD$ is divided into three equal segments.

$\blacksquare$