Problem 1567

1. **Part 1: Show that the line $DM$ passes through the midpoint of $BC$**

Let $D = (0,0)$, $A = (0,2b)$, $C = (2a,0)$, and $B = (2a,2b)$. The midpoint $I$ of $CD$ is $(a,0)$.

The slope of line $BI$ is:
$$\text{slope of } BI = \frac{2b - 0}{2a - a} = \frac{2b}{a}$$
The equation of line $BI$ is:
$$y = \frac{2b}{a}(x - a)$$

Since $AC$ is the diagonal of the rectangle, its equation can be found using the points $A$ and $C$:
$$\text{slope of } AC = \frac{0 - 2b}{2a - 0} = -\frac{b}{a}$$
The equation of line $AC$ is:
$$y - 2b = -\frac{b}{a}(x - 0) \implies y = -\frac{b}{a}x + 2b$$

To find the intersection $M$ of $BI$ and $AC$, set the equations equal:
$$\frac{2b}{a}(x - a) = -\frac{b}{a}x + 2b$$
Simplify and solve for $x$:
$$2bx - 2ba = -bx + 2ab \implies 3bx = 4ab \implies x = \frac{4a}{3}$$
Substitute $x = \frac{4a}{3}$ back into the equation of $AC$:
$$y = -\frac{b}{a} \left( \frac{4a}{3} \right) + 2b = -\frac{4b}{3} + 2b = \frac{2b}{3}$$
Thus, $M \left( \frac{4a}{3}, \frac{2b}{3} \right)$.

The midpoint of $BC$ is:
$$\left( \frac{2a + 2a}{2}, \frac{2b + 0}{2} \right) = (2a, b)$$

The slope of line $DM$ is:
$$\text{slope of } DM = \frac{\frac{2b}{3} - 0}{\frac{4a}{3} - 0} = \frac{2b}{4a} = \frac{b}{2a}$$
The equation of line $DM$ is:
$$y = \frac{b}{2a}x$$
Plugging in $x = 2a$:
$$y = \frac{b}{2a} \cdot 2a = b$$
Therefore, $DM$ passes through the midpoint of $BC$.

2. **Part 2: Show that $EM$ bisects $\angle AMB$**

Since $\triangle AEB$ is a 45-45-90 triangle, $AB = x\sqrt{2}$.

Let $D = (0,0)$, $A = (0,x)$, $C = (x\sqrt{2},0)$, $B = (x\sqrt{2},x)$, and $I = \left( \frac{x\sqrt{2}}{2}, 0 \right)$.

The slope of $AC$ is:
$$\text{slope of } AC = \frac{0 - x}{x\sqrt{2} - 0} = -\frac{1}{\sqrt{2}}$$
The slope of $BI$ is:
$$\text{slope of } BI = \frac{x - 0}{x\sqrt{2} - \frac{x\sqrt{2}}{2}} = \frac{x}{\frac{x\sqrt{2}}{2}} = \sqrt{2}$$
Since the slopes multiply to -1:
$$\left( -\frac{1}{\sqrt{2}} \right) \left( \sqrt{2} \right) = -1$$
Thus, $AC \perp BI$, making $\angle AMB = 90^\circ$.

Since $AEBM$ is a cyclic quadrilateral, $\angle AMB = \angle ABE$ and $\angle EMB = \angle EAB$. By the Base Angle Theorem, $\angle EAB = \angle EBA$, so $EM$ bisects $\angle AMB$.

3. **Part 3: Find the area of $AEBM$ in terms of $x$**

The area of $\triangle AEB$ is:
$$[AEB] = \frac{1}{2} \times x \times x = \frac{x^2}{2}$$

From Part 1, the height from $CI$ to $M$ is half as large as the height from $AB$ to $M$, so the height from $AB$ to $M$ is $\frac{2x}{3}$.

The area of $\triangle ABM$ is:
$$[ABM] = \frac{1}{2} \times x\sqrt{2} \times \frac{2x}{3} = \frac{x^2 \sqrt{2}}{3}$$

The total area of $AEBM$ is:
$$[AEBM] = [AEB] + [ABM] = \frac{x^2}{2} + \frac{x^2 \sqrt{2}}{3} = x^2 \left( \frac{1}{2} + \frac{\sqrt{2}}{3} \right) = \frac{x^2 (3 + 2\sqrt{2})}{6}$$

The final answer is $\boxed{ \frac{x^2 (3 + 2\sqrt{2})}{6} }$