1. **State the problem:** We have a circle centered at point $P$ with diameters $\overline{AD}$ and $\overline{BE}$. We need to find the measure of the minor arc $\stackrel{\large{\frown}}{CD}$ in degrees. 2. **Identify given information:** - $\overline{AD}$ and $\overline{BE}$ are diameters, so points $A, D, B, E$ lie on the circle. - Points $A, B, C, D, E$ lie on the circle in clockwise order. - Angle $A P E$ is a right angle ($90^\circ$). - Angle $D P E = 33K - 9$ degrees. - Angle $C P D = 20K + 4$ degrees. 3. **Analyze angles at the center:** Since $A, B, C, D, E$ lie on the circle, arcs are defined by the central angles: - $\angle A P E = 90^\circ$ corresponds to arc $\stackrel{\large{\frown}}{AE}$. - $\angle D P E = 33K - 9$ degrees corresponds to arc $\stackrel{\large{\frown}}{DE}$. - $\angle C P D = 20K + 4$ degrees corresponds to arc $\stackrel{\large{\frown}}{CD}$ (the quantity we want). 4. **Use diameters properties:** - Since $\overline{AD}$ is a diameter, $A$ and $D$ are opposite points. - Similarly, since $\overline{BE}$ is a diameter, $B$ and $E$ are opposite points. - The full circle is $360^\circ$. 5. **Relate angles and arcs around the circle:** Because $A, B, C, D, E$ lie in clockwise order, the arcs add to $360^\circ$: $$\stackrel{\large{\frown}}{AB} + \stackrel{\large{\frown}}{BC} + \stackrel{\large{\frown}}{CD} + \stackrel{\large{\frown}}{DE} + \stackrel{\large{\frown}}{EA} = 360^\circ$$ Replace arcs with central angles: - Arc $\stackrel{\large{\frown}}{EA}$ corresponds to angle $A P E = 90^\circ$ (given). - Arc $\stackrel{\large{\frown}}{DE}$ corresponds to angle $D P E = 33K - 9$. - Arc $\stackrel{\large{\frown}}{CD}$ corresponds to angle $C P D = 20K + 4$ (unknown to find). Since $A$ and $D$ are endpoints of a diameter, the arc $\stackrel{\large{\frown}}{AD}$ is $180^\circ$. Similarly, $B$ and $E$ are endpoints of a diameter, so arc $\stackrel{\large{\frown}}{BE} = 180^\circ$. The points are ordered $A, B, C, D, E$, so we can split the circle accordingly. 6. **Set up equations for the arcs:** - Arc $\stackrel{\large{\frown}}{AE} = 90^\circ$ - Arc $\stackrel{\large{\frown}}{DE} = 33K - 9$ - Arc $\stackrel{\large{\frown}}{CD} = 20K + 4$ Because $\overline{AD}$ is a diameter, the arc $\stackrel{\large{\frown}}{AD} = 180^\circ$. We also know $\stackrel{\large{\frown}}{AE} + \stackrel{\large{\frown}}{ED} + \stackrel{\large{\frown}}{DA} = 360^\circ$ in some arrangement. 7. **Calculate $K$ using the angle sum around $P$ in triangle formed:** Since $B$ and $E$ are endpoints of diameter $\overline{BE}$, the angle $A P E = 90^\circ$ We understand from the problem $D P E$ and $C P D$ are parts of the angle distribution along the circle. 8. **Sum arcs that span from $A$ around to $E$ via $C$ and $D$: Total circle: $360^\circ = 90 + (33K - 9) + (20K + 4) + $ other arcs (not explicitly given). Since the problem wants only minor arc $\stackrel{\large{\frown}}{CD}$, use: - Recall $\angle DPE = 33K - 9$ - $\angle CPD = 20K + 4$ Sum of these angles form some part of circle, but it is not strictly solvable without assuming $K$. 9. **Assuming the angles $\angle DPE$ and $\angle CPD$ are adjacent central angles in the circle and total together with $\angle APE$ to 180$^\circ$ (as four points on half circle):** $$90 + (33K - 9) + (20K + 4) = 180$$ Simplify: $$90 + 33K - 9 + 20K + 4 = 180$$ $$33K + 20K + (90 - 9 + 4) = 180$$ $$53K + 85 = 180$$ Solving for $K$: $$53K = 180 - 85 = 95$$ $$K = \frac{95}{53}$$ 10. **Find the measure of minor arc $\stackrel{\large{\frown}}{CD}$:** $$\stackrel{\large{\frown}}{CD} = 20K + 4 = 20 \times \frac{95}{53} + 4 = \frac{1900}{53} + 4 = \frac{1900}{53} + \frac{212}{53} = \frac{2112}{53} \approx 39.87^\circ$$ **Final answer:** $$\boxed{39.87^\circ}$$