In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$.
We have the following diagram:

Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$.
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$. We know that $\frac{XW}{AB} = \frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$.
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = \boxed{242}.$
~Cytronical