On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, two points $D$ and $E$ are fixed, respectively, such that $\overline{A D}=\overline{B E}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-16.jpg?height=656&width=831&top_left_y=500&top_left_x=515)

If the segments $A E$ and $C D$ intersect at point $P$, determine $\measuredangle A P C$.

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Solution

Consider the triangles $D A C$ and $E B A$. It is known that $\overline{D A}=\overline{E B}$. Furthermore, since triangle $A B C$ is equilateral, then $\overline{A C}=\overline{B A}$. Additionally, in an equilateral triangle, all internal angles measure $60^{\circ}$. Therefore, $\measuredangle D A C=\measuredangle E B A=60^{\circ}$. This implies that the triangles $D A C$ and $E B A$ are congruent, and thus $\measuredangle D C A=\measuredangle B A E$. In the figure below, we represent $\measuredangle D C A=\measuredangle B A E=x$.

![](https://cdn.mathpix.com/cropped/2024_05_01_cc59d970f26105cc4ca1g-43.jpg?height=460&width=582&top_left_y=684&top_left_x=840)

Now note that $\measuredangle P A C=60^{\circ}-\measuredangle D A P=60^{\circ}-\measuredangle P C A$. That is, $\angle P A C+\measuredangle P C A=60^{\circ}$. Since the sum of the interior angles in triangle $A P C$ must be $180^{\circ}$, we have that $\measuredangle P A C+\measuredangle P C A+\measuredangle A P C=180^{\circ}$. Therefore, we conclude that $\measuredangle A P C=120^{\circ}$.