Problem 1447

1. Assign Coordinates and Calculate Slopes:
- Given points: $A(0,0)$, $C(5,0)$, and $D\left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$.
- Calculate the slope of line $AD$:
$$m_{AD} = \frac{\frac{3\sqrt{3}}{2} - 0}{-\frac{3}{2} - 0} = -\sqrt{3}$$
- This implies that $\angle DAC = 120^\circ$ because the slope $-\sqrt{3}$ corresponds to an angle of $120^\circ$ with the positive x-axis.

2. **Equation of Line $AE$:**
- Since $\angle CAD$ is bisected by $AE$, the slope of $AE$ is the average of the slopes of $AC$ and $AD$.
- Slope of $AC$ is 0 (horizontal line).
- Slope of $AD$ is $-\sqrt{3}$.
- Average slope for the angle bisector:
$$m_{AE} = \frac{0 + (-\sqrt{3})}{2} = -\frac{\sqrt{3}}{2}$$
- Equation of line $AE$:
$$y = -\frac{\sqrt{3}}{2} x$$

3. **Intersection of $AE$ and $CD$:**
- Equation of line $CD$:
$$y - \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \left( x + \frac{3}{2} \right)$$
- Simplify the equation of $CD$:
$$y = \frac{3\sqrt{3}}{2} x + \frac{3\sqrt{3}}{2} \cdot \frac{3}{2} + \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} x + \frac{9\sqrt{3}}{4}$$
- Set $y = -\frac{\sqrt{3}}{2} x$ equal to the equation of $CD$:
$$-\frac{\sqrt{3}}{2} x = \frac{3\sqrt{3}}{2} x + \frac{9\sqrt{3}}{4}$$
- Solve for $x$:
$$-\frac{\sqrt{3}}{2} x - \frac{3\sqrt{3}}{2} x = \frac{9\sqrt{3}}{4}$$
$$-2\sqrt{3} x = \frac{9\sqrt{3}}{4}$$
$$x = -\frac{9}{8}$$
- Substitute $x$ back into $y = -\frac{\sqrt{3}}{2} x$:
$$y = -\frac{\sqrt{3}}{2} \left( -\frac{9}{8} \right) = \frac{9\sqrt{3}}{16}$$
- Point $E \left( -\frac{9}{8}, \frac{9\sqrt{3}}{16} \right)$.

4. **Calculate Lengths $AE$ and $BE$:**
- Length $AE$:
$$AE = \sqrt{\left( -\frac{9}{8} - 0 \right)^2 + \left( \frac{9\sqrt{3}}{16} - 0 \right)^2} = \sqrt{\left( \frac{9}{8} \right)^2 + \left( \frac{9\sqrt{3}}{16} \right)^2}$$
$$AE = \sqrt{\frac{81}{64} + \frac{243}{256}} = \sqrt{\frac{324}{256}} = \frac{9}{8} \sqrt{2}$$
- Length $BE$:
- Using the coordinates of $B \left( -\frac{1}{2}, -\frac{5\sqrt{3}}{2} \right)$:
$$BE = \sqrt{\left( -\frac{1}{2} + \frac{9}{8} \right)^2 + \left( -\frac{5\sqrt{3}}{2} - \frac{9\sqrt{3}}{16} \right)^2}$$
$$BE = \sqrt{\left( \frac{4}{8} \right)^2 + \left( -\frac{40\sqrt{3}}{16} - \frac{9\sqrt{3}}{16} \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + \left( -\frac{49\sqrt{3}}{16} \right)^2}$$
$$BE = \sqrt{\frac{1}{4} + \frac{2401}{256}} = \sqrt{\frac{2405}{256}} = \frac{49}{8}$$

5. **Sum $AE + BE$:**
$$AE + BE = \frac{9}{8} \sqrt{2} + \frac{49}{8} = 8$$