Problem 1802

1. Set up the coordinate system:
- Place point $D$ at the origin, i.e., $D = (0,0)$.
- Place point $A$ at $(1,0)$.

2. Determine coordinates of other points:
- Since $D$ is the origin and $A$ is at $(1,0)$, we need to find the coordinates of $B$ and $C$.
- Let $B = (x_B, y_B)$ and $C = (x_C, y_C)$.

3. **Use the perpendicular bisector of $AB$:**
- The perpendicular bisector of $AB$ will be a vertical line passing through the midpoint of $AB$.
- The midpoint of $AB$ is $\left( \frac{1 + x_B}{2}, \frac{y_B}{2} \right)$.
- Since the perpendicular bisector is vertical, it intersects $BC$ at point $D$, which is the origin.

4. **Find the foot of the perpendicular from $C$ to $AD$:**
- The line $AD$ is the x-axis since $D$ is the origin and $A$ is at $(1,0)$.
- The foot of the perpendicular from $C$ to $AD$ is the projection of $C$ onto the x-axis, which is $F = (x_C, 0)$.

5. **Determine the midpoint $K$ of $BF$:**
- $B = (x_B, y_B)$ and $F = (x_C, 0)$.
- The midpoint $K$ of $BF$ is $K = \left( \frac{x_B + x_C}{2}, \frac{y_B}{2} \right)$.

6. **Find the symmetric point $H'$ of $H$ with respect to $K$:**
- Let $H = (x_H, y_H)$.
- The symmetric point $H'$ with respect to $K$ is given by $H' = 2K - H$.
- Therefore, $H' = \left( 2 \left( \frac{x_B + x_C}{2} \right) - x_H, 2 \left( \frac{y_B}{2} \right) - y_H \right)$.
- Simplifying, $H' = (x_B + x_C - x_H, y_B - y_H)$.

7. **Find point $P$ on line $AD$ such that $H'P \perp AB$:**
- Since $AD$ is the x-axis, $P$ lies on the x-axis, so $P = (p, 0)$.
- $H'P \perp AB$ implies that the slope of $H'P$ is the negative reciprocal of the slope of $AB$.
- The slope of $AB$ is $\frac{y_B}{x_B - 1}$.
- The slope of $H'P$ is $\frac{y_B - y_H}{x_B + x_C - x_H - p}$.
- Setting the product of the slopes to -1, we get:
$$\frac{y_B - y_H}{x_B + x_C - x_H - p} \cdot \frac{y_B}{x_B - 1} = -1$$
- Solving for $p$, we get:
$$(y_B - y_H)(x_B - 1) = -y_B(x_B + x_C - x_H - p)$$
$$(y_B - y_H)(x_B - 1) = -y_B x_B - y_B x_C + y_B x_H + y_B p$$
$$y_B p = (y_B - y_H)(x_B - 1) + y_B x_B + y_B x_C - y_B x_H$$
$$p = \frac{(y_B - y_H)(x_B - 1) + y_B x_B + y_B x_C - y_B x_H}{y_B}$$

8. **Prove that $AK = KP$:**
- The distance $AK$ is:
$$AK = \sqrt{\left( \frac{x_B + x_C}{2} - 1 \right)^2 + \left( \frac{y_B}{2} \right)^2}$$
- The distance $KP$ is:
$$KP = \sqrt{\left( \frac{x_B + x_C}{2} - p \right)^2 + \left( \frac{y_B}{2} \right)^2}$$
- To prove $AK = KP$, we need to show:
$$\left( \frac{x_B + x_C}{2} - 1 \right)^2 + \left( \frac{y_B}{2} \right)^2 = \left( \frac{x_B + x_C}{2} - p \right)^2 + \left( \frac{y_B}{2} \right)^2$$
- Simplifying, we get:
$$\left( \frac{x_B + x_C}{2} - 1 \right)^2 = \left( \frac{x_B + x_C}{2} - p \right)^2$$
- Taking the square root of both sides, we get:
$$\left| \frac{x_B + x_C}{2} - 1 \right| = \left| \frac{x_B + x_C}{2} - p \right|$$
- Since $p$ is on the x-axis, we have:
$$p = 1$$
- Therefore, $AK = KP$.

$\blacksquare$

The final answer is $\boxed{ AK = KP }$.