$ABCD$ is a rectangle. $I$ is the midpoint of $CD$. $BI$ meets $AC$ at $M$. Show that the line $DM$ passes through the midpoint of $BC$. $E$ is a point outside the rectangle such that $AE = BE$ and $\angle AEB = 90^o$. If $BE = BC = x$, show that $EM$ bisects $\angle AMB$. Find the area of $AEBM$ in terms of $x$.
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 \( \boxed{ \frac{x^2 (3 + 2\sqrt{2})}{6} } \)