[Geometry]
Problem: In quadrilateral $A B C D, \mathrm{~m} \angle B+\mathrm{m} \angle D=270^{\circ}$. The circumcircle of $\triangle A B D$ intersects $\overline{C D}$ at point $E$, distinct from $D$. Given that $B C=4, C E=5$, and $D E=7$, compute the diameter of the circumcircle of $\triangle A B D$.

Solution: Note that $\mathrm{m} \angle A+\mathrm{m} \angle C=90^{\circ}$ in quadrilateral $A B C D$. Because quadrilateral $A B E D$ is cyclic, it follows that $\mathrm{m} \angle A D E+\mathrm{m} \angle A B E=180^{\circ}$. Moreover, because $\mathrm{m} \angle A B E+\mathrm{m} \angle E B C+\mathrm{m} \angle A D E=270^{\circ}$, it follows that $\angle E B C$ is a right angle. Thus $B E=\sqrt{C E^{2}-B C^{2}}=\sqrt{5^{2}-4^{2}}=3$. Let $\mathrm{m} \angle B E C=\theta$; then $\cos \theta=\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$.

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Applying the Law of Cosines to $\triangle B E D$ yields

$$
B D^{2}=3^{2}+7^{2}-2 \cdot 3 \cdot 7 \cos \left(180^{\circ}-\theta\right)=3^{2}+7^{2}+2 \cdot 3 \cdot 7 \cos \theta=\frac{416}{5}
$$

Thus $B D=\frac{4 \sqrt{26}}{\sqrt{5}}$. Let $R$ be the circumradius of $\triangle A B D$ and $\triangle B E D$. Then the requested diameter is $2 R$, and



applying the Law of Sines to $\triangle B E D$ yields

$$
2 R=\frac{B D}{\sin \left(180^{\circ}-\theta\right)}=\frac{B D}{\sin \theta}=\frac{4 \sqrt{26}}{\sqrt{5}} \cdot \frac{5}{4}=\sqrt{\mathbf{1 3 0}}
$$

Final answer: $\sqrt{130}$