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!
$\angle CMB = \angle CMA = 90^\circ$, 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 $\angle CMB$.