1. **Problem Statement:** Given a circle with diameter $BE = 10$, chord $CD = 6$ is bisected by $BE$, and arc $CD$ measures $90^\circ$. We need to find the measure of arc $DE$, length of line $FD$, and length of line $AF$. 2. **Known Information:** - Diameter $BE = 10$ implies radius $r = \frac{10}{2} = 5$. - Arc $CD = 90^\circ$. - Chord $CD = 6$. - $BE$ bisects chord $CD$, so $F$ is midpoint of $CD$. 3. **Find the measure of arc $DE$:** - The total circle is $360^\circ$. - Since $BE$ is diameter, it divides the circle into two semicircles of $180^\circ$ each. - Arc $CD = 90^\circ$ lies on one semicircle. - The remaining arc on the same semicircle is $180^\circ - 90^\circ = 90^\circ$. - Therefore, arc $DE = 90^\circ$. 4. **Find length of line $FD$:** - Since $F$ bisects chord $CD$, $CF = FD = \frac{CD}{2} = \frac{6}{2} = 3$. 5. **Find length of line $AF$:** - Point $A$ is not defined in the problem; assuming $A$ is the center of the circle (midpoint of $BE$), which is common in such problems. - Since $F$ lies on chord $CD$, we find distance $AF$ where $A$ is center. 6. **Calculate $AF$:** - The chord $CD$ is bisected by $BE$ at $F$, so $F$ lies on $BE$. - Since $BE$ is diameter, $A$ is midpoint of $BE$, so $A$ is center. - The distance from center $A$ to midpoint $F$ of chord $CD$ can be found using Pythagoras: Chord length $CD = 6$, radius $r = 5$. Half chord length $CF = 3$. Distance from center to chord $d = \sqrt{r^2 - (\frac{CD}{2})^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$. Since $F$ lies on $BE$ (diameter), and $A$ is center, $AF = d = 4$. **Final answers:** - Measure of arc $DE = 90^\circ$. - Length of line $FD = 3$. - Length of line $AF = 4$.