Kupptsov L.

Given an equilateral triangle $ABC$. A certain line, parallel to the line $AC$, intersects lines $AB$ and $BC$ at points $M$ and $P$ respectively. Point $D$ is the center of the equilateral triangle $PMB$, and point $E$ is the midpoint of segment $AP$. Find the angles of triangle $DEC$.
Let $C_{1}$ be the image of point $C$ under symmetry with respect to point $E$. Then triangle $C_{1} D M$ is equal to triangle $C D P$. Therefore, $C_{1} D = C D$.

## Solution

Let $K$ be the projection of point $D$ onto $B C$, and $Q$ be the projection of point $B$ onto $A C$. Points $K, D, Q$, and $C$ lie on a circle with diameter $C D$. Since

$$
\begin{gathered}
\angle E Q D = \angle A Q D - \angle A Q E = 90^{\circ} - 60^{\circ} = 30^{\circ}, \\
\angle D K E = \angle D K C - \angle E K C = 90^{\circ} - \angle A B C = 90^{\circ} - 60^{\circ} = 30^{\circ},
\end{gathered}
$$

the segment $D E$ is seen from points $Q$ and $K$ at the same angle. Therefore, points $K, D, E, Q$ lie on the same circle. Hence,

$$
\angle D E C = 90^{\circ}, \angle D C E = \angle D Q E = 30^{\circ}.
$$

Second method.

Let $C_{1}$ be the image of point $C$ under symmetry with respect to point $E$. Triangle $C_{1} D M$ is equal to triangle $C D P$, because

$$
C_{1} M = C_{1} P - M P = A C - M P = B C - B P = P C, M D = D P, \angle C_{1} M D = \angle C P D = 150^{\circ}
$$

Therefore, $C_{1} D = C D$. Consequently, the median $D E$ of the isosceles triangle $C_{1} D C$ is its height. Let $K$ be the midpoint of segment $B P$. Then $E K$ is the midline of triangle $A P B$, and since points $E$ and $K$ lie on a circle with diameter $C$, we have

$$
\angle E D C = \angle E K C = 60^{\circ}, \angle D E C = 90^{\circ}, \angle D C E = 30^{\circ}
$$

Third method.

Rotate triangle $B M P$ by $60^{\circ}$ around point $C$ so that point $B$ moves to $A$. Then point $P$ will move to point $P_{1}$ on segment $A C$, point $M$ will move to point $M_{1}$ outside triangle $A B C$, and point $D$ will move to point $D_{1}$, the centroid of triangle $A M_{1} P_{1}$.

Quadrilateral $D P D_{1} A$ is a parallelogram, and $D D_{1}$ is its diagonal. Therefore, $D_{1} D$ passes through point $E$ and $D_{1} E = D E$.

Since $C E$ is the median of the isosceles triangle $D C D_{1} (C D = C D_{1})$, we have $\angle C E D = 90^{\circ}$, and since $\angle D C D_{1} = 60^{\circ}$, we have $\angle D C E = 30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-63.jpg?height=435&width=549&top_left_y=1532&top_left_x=366)

Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-63.jpg?height=420&width=644&top_left_y=1522&top_left_x=1050)

Fig. 2.

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-63.jpg?height=484&width=626&top_left_y=1959&top_left_x=796)

## Answer

$90^{\circ}, 60^{\circ}, 30^{\circ}$.