Quadrilateral Side Lengths

We note that triangles $ABC$ and $ACD$ are right triangles. By the Pythagorean theorem, we have $AB = \sqrt{AC^2 - BC^2} = \sqrt{30 ^2 - 18 ^2} = 24$ and $CD = \sqrt{AD^2 - AC^2} = \sqrt{50 ^2 - 30 ^2} = 40$ .

Solution 1: Since $CB:BA:AC = AC:CD:DA=3:4:5$ we have that triangles $ABC$ and $DCA$ are similar by side-side-side.

So $\angle CAD = \angle ACB = 90^\circ - \angle BAC \Rightarrow \angle BAD = \angle BAC + \angle CAD = 90^\circ$ . Thus $BAD$ is a right angle triangle. By the Pythagorean theorem, we have $BD = \sqrt{AB^2 + AD^2} = \sqrt{24 ^2 + 50 ^2} = 2 \sqrt{144 + 625} = 2 \sqrt{769}$ . Hence $a + b = 2 + 769 = 771$ .

Solution 2: We have that $\cos \left(\angle BCD \right) = \cos \left(90^\circ + \angle ACB\right) = -\sin \left(\angle ACB\right) = -\frac{24}{30} = -\frac{4}{5}$ . Applying the cosine law on triangle $BCD$ , we have $\begin{aligned} BD^2 &= BC^2 + CD^2 - 2(BC)(CD)\cos \left(\angle BCD\right) \\ &= 18 ^2 + 40 ^2 + 2(18)(40) \left(\frac{4}{5}\right) \\ &= 3076 \\ \end{aligned}$

Thus $BD = 2\sqrt{769}$ . Hence $a + b = 2 + 769 = 771$ .