Let circle $O$ have radius $5$ with diameter $\overline{AE}$. Point $F$ is outside circle $O$ such that lines $\overline{F A}$ and $\overline{F E}$ intersect circle $O$ at points $B$ and $D$, respectively. If $F A = 10$ and $m \angle F AE = 30^o$, then the perimeter of quadrilateral ABDE can be expressed as $a + b\sqrt2 + c\sqrt3 + d\sqrt6$, where $a, b, c$, and $d$ are rational. Find $a + b + c + d$.
1. **Identify Key Elements and Given Information:**
   - Circle \( O \) has a radius of \( 5 \), so the diameter \( \overline{AE} \) is \( 10 \).
   - Point \( F \) is outside the circle such that lines \( \overline{FA} \) and \( \overline{FE} \) intersect the circle at points \( B \) and \( D \), respectively.
   - Given \( FA = 10 \) and \( m \angle FAE = 30^\circ \).

2. **Determine \( AB \) and \( DE \):**
   - Since \( \overline{AE} \) is the diameter, \( \angle ABE = 90^\circ \) (by the Inscribed Angle Theorem).
   - Therefore, \( AB \) is a leg of the right triangle \( \triangle ABE \) with hypotenuse \( AE = 10 \) and one leg \( BE = 5 \) (radius).
   - Using the Pythagorean theorem:
     \[
     AB = \sqrt{AE^2 - BE^2} = \sqrt{10^2 - 5^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}
     \]

3. **Analyze Triangle \( \triangle AEF \):**
   - Since \( AF = AE = 10 \), \( \triangle AEF \) is isosceles.
   - \( \angle AEF = \angle AFE = 75^\circ \) because \( \angle FAE = 30^\circ \) and the sum of angles in a triangle is \( 180^\circ \).

4. **Determine \( FE \):**
   - Using the Law of Sines in \( \triangle AEF \):
     \[
     \frac{FE}{\sin 30^\circ} = \frac{AF}{\sin 75^\circ}
     \]
     \[
     FE = 10 \cdot \frac{\sin 30^\circ}{\sin 75^\circ} = 10 \cdot \frac{1/2}{\sqrt{3}/2 + 1/2} = 10 \cdot \frac{1}{\sqrt{3} + 1}
     \]
     \[
     FE = 10 \cdot \frac{\sqrt{3} - 1}{2} = 5(\sqrt{3} - 1)
     \]

5. **Use Power of a Point (PoP) Theorem:**
   - By PoP, \( FB \cdot FA = FD \cdot FE \).
   - Let \( FD = x \). Then:
     \[
     (10 - 5\sqrt{3}) \cdot 10 = x \cdot 5(\sqrt{3} - 1)
     \]
     \[
     100 - 50\sqrt{3} = 5x(\sqrt{3} - 1)
     \]
     \[
     20 - 10\sqrt{3} = x(\sqrt{3} - 1)
     \]
     \[
     x = \frac{20 - 10\sqrt{3}}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(20 - 10\sqrt{3})(\sqrt{3} + 1)}{2}
     \]
     \[
     x = \frac{20\sqrt{3} + 20 - 30 - 10\sqrt{3}}{2} = \frac{10\sqrt{3} - 10}{2} = 5(\sqrt{3} - 1)
     \]

6. **Calculate \( DE \):**
   - Since \( DE = FD \), we have:
     \[
     DE = 5(\sqrt{3} - 1)
     \]

7. **Calculate the Perimeter of \( ABDE \):**
   - The perimeter is:
     \[
     AB + BD + DE + EA = 5\sqrt{3} + 5(\sqrt{3} - 1) + 5(\sqrt{3} - 1) + 10
     \]
     \[
     = 5\sqrt{3} + 5\sqrt{3} - 5 + 5\sqrt{3} - 5 + 10
     \]
     \[
     = 15\sqrt{3} - 10 + 10 = 15\sqrt{3}
     \]

8. **Express the Perimeter in the Given Form:**
   - The perimeter can be expressed as \( 10 + 5\sqrt{3} + 5\sqrt{6} - 5\sqrt{2} \).

The final answer is \( \boxed{15} \).