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#1 h Aug 17, 2011, 1:22 AM • 2 Y

Y by Adventure10, Mango247

In the convex pentagon ABCDE, the sides BC, CD and DE are equal and each diagonal is parallel to a side.
Prove that ABCDE is a regular pentagon.

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#2 h Aug 17, 2011, 4:06 AM • 3 Y

Y by math07, Adventure10, Mango247

Consider side $AB$ of pentagon $ABCDE.$ Diagonal $CE$ is the only diagonal that does not contain either $A$ or $B.$ Thus, for any side, only one diagonal could possibly be parallel to it.

Now, consider the three given sides, $BC=CD=DE.$ Without loss of generality, let them all be $1.$ Since $BE \| CD,$ let $\angle BCD = \angle EDC = \alpha.$ Let $k$ be the line through $D$ such that $k \| BC.$ Let $\ell$ be the line through $C$ such that $\ell \| DE.$ Let $m$ be the line through $E$ such that $m \| BD.$ Let $n$ be the line through $B$ such that $n \| CE.$ We must prove that $k \cap \ell = m \cap n.$

Since $BCDE$ is an isosceles trapezoid, we have that $BE= 2\sin \left(\alpha-\frac{\pi}{2}\right) + 1$ and that the distance between $BE$ and $CD$ (let this value be $x$ ) is $\cos \left(\alpha - \frac{\pi}{2} \right).$ Look at the altitude from $A$ to $CD,$ we have the equation

$\frac{1}{2}BE \cdot \tan \left( \frac{\pi - \alpha}{2} \right) + x = \frac{1}{2} \cdot \tan (\pi - \alpha)$

Plugging in the values of $BE$ and $x,$ we find that $x=108^\circ ,$ which implies that the given pentagon is regular.

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