euclidean-geometrygeometrytriangles
A $\triangle ABC$ is drawn ($\angle C = 90^\circ$), in which $CL$ $(L
\in AB)$ is bisector. The circle $k$ with diameter $CL$ intersects AB,
BC and CA, respectively, in $M$, $N$ and $P$. Show that $MN$ and $MP$
are angle bisectors of $\angle BMC$ and $\angle AMC$.
$PMNC$ is a cyclic quadrilateral, thus $\angle PMN + \angle PCN = 180^\circ$ and $\angle PMN = 90^\circ$. Also $\angle CML = 90^\circ$. I don't know what to do after this. I would be very grateful if you could help me!
I'll describe my methods:
1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.
2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).
Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.
3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$
Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.
At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $ ∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.
4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.
Extend $DQ$ to meet the circle at $X$.Then, BX would be parallel to the simson line as mentioned in the comment.
If part: Now,let $BX$ meet $AC$ at $Y$.Take perspectivity at $B$ onto line $AC$ to show that $(Y,Q,A,C)$ harmonic.Now,take perspectivity at $X$ onto the circle to show that $ABCD$ is harmonic. The other direction can be proved similarly.
Best Answer
$\angle CMB = \angle CMA = 90^\circ$, since $\angle CML$ subtends the diameter $CL$.
By the inscribed angle theorem, $\angle LPC=90^\circ = \angle LNC$
Triangles $LPC$ and $LNC$ are equal (by common side and equal angles), so $CP = CN$ and $CPN$ is an isosceles right triangle.
By the inscribed angle theorem, $\angle CMP=\angle CNP=45^\circ$, so $MP$ is the angle bisector of $\angle CMA$.
Analogically, $MN$ is the angle bisector of $\angle CMB$.