The segment $AB$ of length $16 \mathrm{~cm}$ is the diameter of a circle with center $O$. A secant line intersects the circle at $C$ and $D$ and the line $AB$ at $P$, as indicated in the following figure. If $OD = DP$ and $\angle APC = 18^{\circ}$, what is the value of the angle $\angle AOC$?

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-09.jpg?height=421&width=600&top_left_y=572&top_left_x=802)
Solution

Since $O D=D P$, it follows that $\angle D O P=\angle O P D=18^{\circ}$. By the Exterior Angle Theorem, it follows that $\angle O D C=2 \cdot 18^{\circ}=36^{\circ}$. Triangle $C O D$ is isosceles, as $C O$ and $O D$ are radii of the circle. Thus, $\angle O C D=\angle C D O=36^{\circ}$. Again, by the Exterior Angle Theorem, it follows that $\angle C O A=\angle O C P+\angle C P O=54^{\circ}$.