Let $A B C D E$ be a convex pentagon. If $\alpha=\angle D A C, \beta=\angle E B D, \gamma=A C E, \delta=\angle B D A$, and $\epsilon=\angle B E C$, as shown in the picture, calculate the sum $\alpha+\beta+\gamma+\delta+\epsilon$.
Let $M, N, P, Q, R$ denote the intersections of the lines $A C$ and $B E, A C$ and $B D, C E$ and $B D, D A$ and $C E, E B$ and $A E$, respectively. Since the sum of the internal angles of a triangle is $180^{\circ}$, from $\triangle A M R$ we get $\alpha=180^{\circ}-\angle A M R-\angle M R A$. We also know that $\angle A M R=$ $\angle B M N$ and we can denote these angles by $\angle M$. Analogously $\angle A R M=\angle E R Q=\angle R$. With similar equations for $\beta, \gamma, \delta$, and $\epsilon$ we get that $\alpha+\beta+\gamma+\delta+\epsilon=5 \cdot 180^{\circ}-$ $2(\angle M+\angle N+\angle P+\angle Q+\angle R)$. Since $\angle M, \angle N, \angle P, \angle Q$, and $\angle R$ are exterior angles of the pentagon $M N P Q R$ their sum has to be $360^{\circ}$ implying that $\alpha+\beta+\gamma+\delta+\epsilon=180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_d90eaa74d4107324eb71g-1.jpg?height=477&width=479&top_left_y=641&top_left_x=1254)