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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!

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    Check angles of the petagon – 2017-02-09
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    @labbhattacharjee Like with quadrilaterals, it is not enough to know that the angles are all equal, however. – 2017-02-09
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    What do you mean by triangles being "equal"? This is not defined. Do you mean congruent? If so, the result is easy. Do you mean equal area? In that case the answer is probably no, but an interesting question to work out. – 2017-02-09
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    @Arthur, But wiki says : "In Euclidean geometry, a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length). " – 2017-02-09
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    @labbhattacharjee That's what I'm saying: equiangular is not enough; you need equilateral as well. – 2017-02-09
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    @Arthur, Thanks for your explanation. – 2017-02-09
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    In this case I suppose that equal means "same angles and same area". – 2017-02-09
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    We obtain five triangles only if the pentagon doesn't have two parallel sides. – 2017-02-09

2 Answers 2

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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.

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    I have understood all, thanks for your answer! – 2017-02-10
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Hint:

AS suggested in the comment, look at the figure and prove that $\beta=\alpha$.

enter image description here

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    And what is $\alpha$? I said, if I have a convex pentagon the center is the barycenter? – 2017-02-09
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    Yes, $\alpha$ is in the barycenter that is the center of the circumscribed circle.And if the pentagon is regular $\alpha=360°/5$. – 2017-02-09
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    Who said the convex pentagon was cyclic? – 2017-02-09