$I$ is a point of the circle of diameter $JK$. The perpendicular bisector of $JK$ cut the semi-circle not containing $I$ at $M$. Let $N$ and $P$ be the orthogonal projections of $M$ on $IJ$ and $IP$. Prove that $\angle JMN=\angle PMK$. The next question asks to prove that $JMN$ and $KMP$ are congruent.
How I did it:
$\angle JMN=90-\angle MJN=90-\left(\angle MJK+\angle KJI\right)=90-(45+\angle KJI)=90-\angle KJI-45=\angle IKJ -45=180-45-\angle MKP -45=90-\angle MKP=\angle PMK$
Maybe there's a shorter way in this case but how can I simplify angle chasing for students?