\section*{Problem 15}

\(\mathrm{ABC}\) is equilateral. \(\mathrm{A}\) line parallel to \(\mathrm{AC}\) meets \(\mathrm{AB}\) at \(\mathrm{M}\) and \(\mathrm{BC}\) at \(\mathrm{P}\). \(\mathrm{D}\) is the center of the equilateral triangle BMP. E is the midpoint of AP. Find the angles of DEC.

\section*{Answer}

\(\mathrm{D}=60^{\circ}, \mathrm{E}=90^{\circ}\)


\section*{Solution}

![](https://cdn.mathpix.com/cropped/2024_05_09_2c0a57c089c74a6cc642g-087.jpg?height=397&width=443&top_left_y=932&top_left_x=247)

Let \(\mathrm{K}\) be the midpoint of \(\mathrm{BP}\) and \(\mathrm{L}\) the midpoint of \(\mathrm{AC}\). EL is parallel to \(\mathrm{BC}\), so \(\angle \mathrm{ELC}=\) \(120^{\circ}\). EK is parallel to \(\mathrm{AB}\), so \(\angle \mathrm{EKC}=60^{\circ}\), so \(\mathrm{ELCK}\) is cyclic. But \(\angle \mathrm{DKC}=\angle \mathrm{DLC}=90^{\circ}\), so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence \(\angle \mathrm{DEC}=\angle \mathrm{DLC}=90^{\circ}\), and \(\angle \mathrm{EDC}=\angle \mathrm{EKC}=60^{\circ}\).

