Regular pentagon If we extend the sides of a convex pentagon we'll obtain 5 triangles. If all the triangles are equal, is a regular pentagon?
I think is true but I'm not sure how to prove it.
Thanks!
 A: Construct five triangles, all having the same angles $\alpha$, $\beta$, $\gamma$, on the sides of the pentagon (assuming such triangles exist and are all external to the pentagon). At every vertex of the pentagon you have two couples of vertical angles: two of them from the triangles and the other two both equal to one of the interior angles of the pentagon. The sum of angles at each vertex is $360°$ and the sum of the interior angles of the pentagon is $540°$, hence the sum of all $10$ triangle angles appearing at a vertex is $720°$. If we take from these only five, one from each couple of vertical angles (we'll call them "vertex angles"), we get that their sum is $360°$.
Suppose all three angles $\alpha$, $\beta$, $\gamma$ are different: as consecutive vertices must have different triangle angles, no angle can be repeated thrice among vertex angles. The only possibility is then a pattern such as $\alpha\alpha\beta\beta\gamma$ for vertex angles and we would have $\alpha+\alpha+\beta+\beta+\gamma=360°$, that is $\alpha+\beta=180°$, which is impossible.
It follows that all triangles must be isosceles, with angles $\alpha$, $\alpha$, $\beta$. Possible vertex patterns could be $\alpha\alpha\alpha\beta\beta$, $\alpha\alpha\alpha\alpha\beta$ and $\alpha\alpha\alpha\alpha\alpha$, but the first two cases would lead to $\alpha+\beta=180°$ and $2\alpha=180°$ and must be discarded. The only possibility is then that all vertex angles $\alpha$ are equal to $72°$, from which it follows that all pentagon angles are equal to $108°$.
All triangles are isosceles with the same base angles: having the same area, they must also have the same base. All the sides of the pentagon are then equal and it is a regular pentagon.
A: Hint:
AS suggested in the comment, look at the figure and prove that $\beta=\alpha$.

