4. Given an acute-angled triangle $A B C$. The feet of the altitudes $B M$ and $C N$ have perpendiculars $M L$ to $N C$ and $N K$ to $B M$. Find the angle at vertex $A$, if the ratio $K L: B C=3: 4$.
# Solution.

We will prove that triangles $N A M$ and $C A B$ are similar with a similarity coefficient of $\cos A$. The ratio of the leg $A M$ to the hypotenuse $A B$ is $\cos A$. Similarly, $A M / A C=\cos A$.

Since angle $A$ is common to both triangles

![](https://cdn.mathpix.com/cropped/2024_05_06_9aa6cec861d5e2a17909g-4.jpg?height=908&width=696&top_left_y=1208&top_left_x=1154)
and the corresponding sides are proportional, the triangles are similar with $k=\cos A$. Therefore, the proportion also holds for the third pair of sides of the similar triangles

$$
\frac{M N}{B C}=\cos A
$$

In quadrilateral ANPM, two opposite right angles exist, so $\angle N P M=180^{\circ}-\angle A \Rightarrow \angle M P L=\angle A \Rightarrow \angle K M L=\angle P M L=90^{\circ}-\angle A$

Two right angles subtend the same hypotenuse, hence they are inscribed in the same circle with radius $R=M N / 2$. By the Law of Sines, the following equalities hold

$$
\frac{K L}{\sin \left(90^{\circ}-A\right)}=2 R=2 N O=M N
$$

$$
\frac{K L}{\cos A}=M N=B C \cos A \Rightarrow \frac{K L}{B C}=\cos ^{2} A \Rightarrow \cos ^{2} A=\frac{3}{4} \Rightarrow \cos A=\frac{\sqrt{3}}{2}
$$

Therefore, $\mathrm{A}=30^{\circ}$

Answer: $\mathrm{A}=30^{\circ}$

## Recommendations for checking.

Do not deduct points for the absence of a proof of the similarity of triangles NAM and CAB. Consider it a known fact.

Do not add points for the presence of only a proof of the similarity of triangles NAM and CAB, i.e., 0 points.

If only the equality of angles $\angle M P L=\angle A$ or $\angle K M L=\angle P M L=$ $90^{\circ}-\angle A$ is proven, 1 point is awarded.

If the circle circumscribed around quadrilateral ANPM is present, 1 point is added.