You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. If $KL$ is the diameter, $KL\parallel AB$, $KM\perp AB$ and $LN\perp AB$ Find angle $\angle MCN$.
Measuring this angle it works out to be $45^o$, but I don't know how to work it out. I attempted to utilize $KLNM$ being a rectangle (something which is proved since $KM\perp AB$ and $LN\perp AB$ which means $KM\parallel LN$ and also $KL\parallel MN$ hence $KLMN \#$ and since $KM\perp MN$ then $KMNL$ is a rectangle), but this didn't work out for me. Could you please explain to me how to solve the question?
Best Answer
It's apparent that $|OM|=|ON|=|OC|(=\sqrt 2\cdot|OK|)$ and that $\angle MON =90^\circ$. Then by the inscribed angle theorem, $\angle MCN =45^\circ$.