9. Pentagon $A B C D E$ is given with the following conditions:
(a) $\angle C B D+\angle D A E=\angle B A D=45^{\circ}, \angle B C D+\angle D E A=300^{\circ}$
(b) $\frac{B A}{D A}=\frac{2 \sqrt{2}}{3}, C D=\frac{7 \sqrt{5}}{3}$, and $D E=\frac{15 \sqrt{2}}{4}$
(c) $A D^{2} \cdot B C=A B \cdot A E \cdot B D$
Compute $B D$.
Answer: $\sqrt{\sqrt{39}}$ As a preliminary, we may compute that by the law of cosines, the ratio $\frac{A D}{B D}=\frac{3}{\sqrt{5}}$.
Now, construct the point $P$ in triangle $A B D$ such that $\triangle A P B \sim \triangle A E D$. Observe that $\frac{A P}{A D}=$ $\frac{A E \cdot A B}{A D \cdot A D}=\frac{B C}{B D}$ (where we have used first the similarity and then condition 3). Furthermore, $\angle C B D=$ $\angle D A B-\angle D A E=\angle D A B-\angle P A B=\angle P A D$ so by SAS, we have that $\triangle C B D \sim \triangle P A D$.
Therefore, by the similar triangles, we may compute $P B=D E \cdot \frac{A B}{A D}=5$ and $P D=C D \cdot \frac{A D}{B D}=7$. Furthermore, $\angle B P D=360-\angle B P A-\angle D P A=360-\angle B C D-\angle D E A=60$ and therefore, by the law of cosines, we have that $B D=\sqrt{39}$.