Geometry question, prove that $\angle APB = \frac12 (\angle AMB + \angle CMD)$

Question

I got the following question:

Prove that $\angle APB = \frac12 (\angle AMB + \angle CMD)$, with the following figure given:

Also, the following information is given:

• $M$ is the centre of the circle

I have tried several things, but after an hour of staring I just couldn't find the answer. I only got the following:

$$\angle ACB = \angle ADB \ \ (\text{inscribed angle})$$ $$\angle APD = \angle BPC \ \ (\text{vertical angle})$$ $$\Delta APD \sim \Delta BPC \ \ (\text{AAA})$$

But after this I got stuck.

Answer 1

Using the inscribed angle theorem, $$\begin{align}\angle{APB}&=180^\circ-(\angle{PAB}+\angle{PBA})\quad(\leftarrow \text{from $\triangle{APB}$})\\&=180^\circ-(180^\circ-\angle{BDA}-\angle{PAD})\quad(\leftarrow \text{from $\triangle{ADB}$})\\&=\angle{BDA}+\angle{PAD}\\&=\angle{BDA}+\angle{CAD}\\&=\frac{1}{2}\angle{AMB}+\frac 12\angle{CMD}\\&=\frac{1}{2}(\angle{AMB}+\angle{CMD})\end{align}$$