Have You Heard Of Symmedian?

I draw a lot more. Let $K$ be the midpoint of $AD$ . $\angle OKP = \angle OBP = 90^\circ$ , which means $OKBP$ is concylic, so $\angle BKP = \angle BOP = \angle BAC$ . Moreover, $\angle BDK = \angle BCA$ , so we have $\Delta BKD$ and $\Delta BAC$ are similar. 'Cause $K$ and $M$ are midpoints of $AD$ and $BC$ , respectively, we also have $\Delta BAD$ and $\Delta MAC$ are similar, which means $\angle BAD = \angle MAC$ , or $AD$ is the symmedian of triangle $ABC$ .

Now, look at the diagram again. Triangle $ABC$ with tangents at $B$ and $C$ cutting at $P$ , then $AP$ is symmedian. What about triangle $DBC$ ? It has tangents at $B$ and $C$ also cutting at $P$ , then $DP$ is also symmedian. That small thing gives us $\angle BDA = \angle MDC$ , or ${\color{#D61F06}\angle \color{#D61F06}B\color{#D61F06}D\color{#D61F06}M \color{#D61F06}= \color{#D61F06}\angle \color{#D61F06}A\color{#D61F06}D\color{#D61F06}C} = \dfrac12\angle AOC$ . Hence, $\angle AOC =54^\circ$ .

Another way to solve the problem, if you don't realize $DA$ is the symmedian of triangle $DBC$ , is by using the point $E$ , which is the intersection of $AM$ and the circle. $\angle BAD = \angle EAC$ means $\stackrel \frown{BD} = \stackrel \frown{EC} \Rightarrow BCED$ is a trapezium. Hence, triangles $BDM$ and $CEM$ are congruent, which also gives us ${\color{#D61F06}\angle \color{#D61F06}B\color{#D61F06}D\color{#D61F06}M \color{#D61F06}= \angle CEM = \color{#D61F06}\angle \color{#D61F06}A\color{#D61F06}D\color{#D61F06}C}$ (match with the upper red part).

Note: After reading the solution, you may find out that $AB$ is tangent to the circumcircle of $BDM$ . In fact, this problem is not original. What it asks is to prove the upper thing.

General formula: the measure of the angle AOC is equal to double the measure of the angle BDM.