4. Point $O$ is the center of the circumscribed circle of an acute-angled triangle $ABC$, and $H$ is the intersection point of its altitudes. It turns out that the line $OH$ is parallel to side $BC$. On the plane, a point $K$ is marked such that $ABHK$ is a parallelogram. Segments $OK$ and $AC$ intersect at point $L$. In what ratio does the perpendicular dropped from point $L$ to segment $AH$ divide $AH?$
Answer: $1: 1$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-03.jpg?height=1040&width=920&top_left_y=505&top_left_x=588)

First solution. Let $D$ be the foot of the altitude from point $A$, and point $E$ be the intersection of this altitude with the circumcircle of triangle $ABC$, point $A'$ be diametrically opposite to point $A$ on this circle, and point $N$ be the second intersection of line $AK$ with this circle. From the parallelism of lines $OH$ and $BC$, it follows that line $OH$ is perpendicular to the altitude $AD$. Since $AA'$ is the diameter of the circle, $\angle AEA' = 90^\circ$ and, therefore, lines $A'E$ and $OH$ are parallel. Thus, $OH$ is the midline of triangle $AA'E$, so $AH = HE$. Further,

$$
\angle CBE = \angle CAE = 90^\circ - \angle ACB = \angle CBH,
$$

so in triangle $BEH$, segment $BD$ is both a bisector and an altitude, and therefore also a median. Thus, $HD = DE$. From the equalities $AH = HE$ and $HD = DE$, we get that $AH: AD = 2: 3$.

By the condition, lines $AK$ and $BH$ are parallel, and line $BH$ is perpendicular to line $AC$, so $\angle NAC = 90^\circ$ and points $C$ and $N$ are diametrically opposite. Therefore, $\angle NBC = 90^\circ$ and lines $NB$ and $AH$ are parallel. Thus, quadrilateral $AHBN$ is a parallelogram. Therefore, $AN = BH = AK$ and segment $CA$ is a median in triangle $KCN$. But segment $KO$ is also a median in this triangle. Therefore, $L$ is the intersection of the medians of this triangle and $AL: AC = 1: 3$. Then, by Thales' theorem, $AM: AD = 1: 3$. But we already know that $AH: AD = 2: 3$, so $AM = MH$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-04.jpg?height=853&width=903&top_left_y=156&top_left_x=588)

Second solution. Note that if points $A', B'$, and $C'$ are the midpoints of sides $BC, CA$, and $AB$ of triangle $ABC$, then triangle $A'B'C'$ is similar to triangle $ABC$ with a similarity coefficient of $\frac{1}{2}$. On the other hand, the center of the circumcircle of triangle $ABC$ is the orthocenter of triangle $A'B'C'$. Therefore, $AH = 2OA'$ and $BH = 2OB'$.

Let $D$ be the foot of the altitude from point $A$, and $M$ be the foot of the perpendicular dropped from point $L$ to $AH$. Line $OA'$ is the perpendicular bisector of segment $BC$, so it is parallel to the altitude $AD$. By the condition, lines $OH$ and $BC$ are parallel, so $OHDA'$ is a rectangle and $HD = OA' = \frac{1}{2}AH$.

In parallelogram $ABHK$, opposite sides are equal, so $AK = BH = 2OB'$. Triangles $ALK$ and $B'LO$ are similar by two angles ($\angle ALK = \angle B'LO$ as vertical angles, $\angle OB'L = 90^\circ = \angle KAL$) and their similarity coefficient is 2. Let $LB' = x$, then $AL = 2x$ and $CB' = AB' = 3x$, since $B'$ is the midpoint of side $AC$. Therefore, $AL: AC = 2x: 6x = 1: 3$ and $AM: AD = 1: 3$, since triangles $ALM$ and $ACD$ are similar. Let $HD = y$, then $OA' = y, AH = 2y$ and $AD = AH + HD = 3y$. Therefore, $AM = y$ and $MH = AH - AM = 2y - y = y$. Thus, $AM: MH = 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-04.jpg?height=851&width=905&top_left_y=1847&top_left_x=587)

Third solution. Note that if points $A', B'$, and $C'$ are the midpoints of sides $BC, CA$, and $AB$ of triangle $ABC$, then triangle $A'B'C'$ is similar to triangle $ABC$ with a similarity coefficient of $\frac{1}{2}$. On the other hand, the center of the circumcircle of triangle $ABC$ is the orthocenter of triangle $A'B'C'$. Therefore, $AH = 2OA'$ and $BH = 2OB'$.

Let $D$ be the foot of the altitude from point $A$, $M$ be the foot of the perpendicular dropped from point $L$ to $AH$, and $P$ be the midpoint of segment $AK$. Line $OA'$ is the perpendicular bisector of segment $BC$, so it is parallel to the altitude $AD$. By the condition, lines $OH$ and $BC$ are parallel, so $OHDA'$ is a rectangle and $HD = OA' = \frac{1}{2}AH$. Thus, $\frac{AH}{AD} = \frac{2}{3}$. By the condition, lines $AK$ and $BH$ are parallel, and line $BH$ is perpendicular to line $AC$, so $\angle KAC = 90^\circ$. By the condition, $ABHK$ is a parallelogram, so $AK = BH$. Segment $B'P$ is the midline of triangle $CAK$, so $B'P = \frac{1}{2}AK = \frac{1}{2}BH = OB'$. Moreover, $OB'$ and $PB'$ are perpendicular to $AC$, so points $O, B'$, and $P$ lie on the same line. Thus, $OP = 2OB' = AK$ and $OP$ is parallel to $AK$. Therefore, $AOPK$ is a parallelogram. Let $Q$ be the intersection of its diagonals, then $AQ = QP$. Therefore, $AB'$ and $OQ$ are medians of triangle $AOP$, and $L$ is their intersection, so $AL: AB' = 2: 3$ and, therefore, $AL: AC = 1: 3$. From the similarity of triangles $AML$ and $ADC$, it follows that $\frac{AM}{AD} = \frac{AL}{AC} = \frac{1}{3}$. Then if $AM = x$, then $AD = 3x$ and $AH = 2x$, so $MH = x$ and $AM: MH = 1: 1$.