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The extensions of the height $C H$ and the bisector $C L$ of triangle $A B C$ intersect the circumcircle of the triangle at points $P$ and $M$ respectively, such that $C P=2 C H, C M=\frac{9}{4} C L$. Find the angles of triangle $A B C$.
From the condition of the problem, it follows that the line $AB$ is the perpendicular bisector of the chord $CP$, which means that $AB$ is the diameter of the circle. Therefore, $\angle ACB = 90^{\circ}$. Let $\angle BAC = \alpha$. Suppose $CL = 4t$. Then $ML = 5t$, and $\angle ABC = 90^{\circ} - \alpha$. Using the Law of Sines in triangles $ACL$ and $BCL$, we find that

$$
\begin{gathered}
AL = \frac{CL \sin \angle CAL}{\sin \angle ACL} = \frac{4t \sin 45^{\circ}}{\sin \alpha} = \frac{4t \sqrt{2}}{2 \sin \alpha} = \frac{2t \sqrt{2}}{\sin \alpha}, \\
BL = \frac{CL \sin 45^{\circ}}{\sin \angle CBL} = \frac{4t \sin 45^{\circ}}{\sin (90^{\circ} - \alpha)} = \frac{4t \sqrt{2}}{2 \cos \alpha} = \frac{2t \sqrt{2}}{\cos \alpha}
\end{gathered}
$$

By the theorem of the product of segments of intersecting chords, $AL \cdot LB = CL \cdot LM$, or

$$
\frac{2t \sqrt{2}}{\sin \alpha} \cdot \frac{2t \sqrt{2}}{\cos \alpha} = 4t \cdot 5t, \frac{4t^2}{\sin \alpha \cos \alpha} = 20t^2, \sin 2\alpha = \frac{4}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b86fbfb53af157d03a0ag-51.jpg?height=1426&width=1452&top_left_y=642&top_left_x=340)

## Answer

$90^{\circ}, \frac{1}{2} \arcsin \frac{4}{5} = \arcsin \frac{1}{\sqrt{5}}, 90^{\circ} - \arcsin \frac{1}{\sqrt{5}}$.