You have triangle $\triangle ABC$ which is orthogonal $\angle C=90^o$ with circle with center O inscribed in it. Find angle $\angle MCN$

Question

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?

Answer 1

It's apparent that $|OM|=|ON|=|OC|(=\sqrt 2\cdot|OK|)$ and that $\angle MON =90^\circ$. Then by the inscribed angle theorem, $\angle MCN =45^\circ$.

Answer 2

Produce line $MK$ to intersect $AC$ at $P$, and produce $NL$ to intersect $BC$ at $Q$. Triangles $MPC$ and $NQC$ are isosceles. Drop altitude $CH$ to base $AB$ to conclude the proof.

Answer 3

If $r$ is the inradius, please note that $CS = KM = r, CT = LN = r$

Also note that $PS = PK$ and $QT = QL$

So we have $PC = PM, QC = QN$

That makes $\triangle PCM$ and $\triangle QCN$ isosceles triangles.

As $\angle BQN = \angle A, \angle BCN = \frac{\angle A}{2}$

And as $\angle APM = \angle B, \angle ACM = \frac{\angle B}{2}$

$\therefore \angle MCN = 45^0$.