Anybody? Any idea?

Carlez Tevos wrote: In convex hexagon $ ABCDEF$ all sides have equal length and $ \angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Prove that the diagonals $ AD,BE,CF$ are concurrent. It isn't difficult to show that $ \angle A=\angle D, \angle B=\angle E,\angle C=\angle F$ This problem is related to the prolem which is referred in International Olympiad 2005, Mexico.

If it really isn't difficult to show, please show it.

See the topic: http://www.mathlinks.ro/viewtopic.php?p=1129352#1129352 I hope you will find the solution soon.

mr.danh wrote: See the topic: http://www.mathlinks.ro/viewtopic.php?p=1129352#1129352 I hope you will find the solution soon. It's a different problem! Please show my problem.

Prove that ABCDEF is a quadrilateral with an inscribed circle. By definition, a tangential quadrilateral has its diagonals $ AB, BE, CF $ concurrent.

Umm.. I guess it's not necessarily a tangential quadrilateral.. It was just a quick thought.

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Consider the following isometry: translate the entire hexagon until $C$ is mapped to $C'=A$, then rotate about $C'=A$ until the image of $D$ coincides with $F$. Notice now that, since $\angle B+\angle D+\angle F=2\pi$, we have $\angle EFE'=2\pi -\angle AFE - \angle AFE'=2\pi-\angle AFE-\angle CDE=\angle ABC$. Thus, $EE'=AC$. Also, by construction, $AE'=CE$. Thus, $CAE'E$ is a parallelogram, so $\angle AE'E=\angle ACE.$ We now seek to prove that $\angle BCD=\angle AFE$, and analogously will follow $\angle B=\angle E, \angle D=\angle A$. In the following angle chase we use the fact that $\triangle ABC$ and $\triangle CDE$ are isosceles to establish $\angle BAC=\angle BCA$ and $\angle CED=\angle DCE$. We also know that the isometry preserves angles between corresponding sides, so $\angle AE'F=\angle C'E'D'=\angle CED$. Since $\triangle EFE'\cong \triangle CBA$, $\angle EFE'=\angle BAC$. Now notice that $F$ is the circumcenter of $AEE'$ since $FA=FE=DE=FE'$. Thus, $\angle AE'E=\frac{1}{2}\angle AFE=\angle AE'F+\angle FE'E=\angle CED+\angle BAC=\angle ECD+\angle BCA=\angle BCD-\angle ACE=\angle BCD-\angle AE'E\iff$ $$\angle AFE=2\angle AE'E=\angle BCD.$$ Thus, opposite pairs of angles are equal. From here, it is easy to show that opposite sides are parallel. Now, notice that since $AF=CD$, and from $\triangle BAC\cong \triangle EDC$ we must have $AC=FD$, $ACDF$ must be a parallelogram. It follows that the diagonals $AD,CF$ and $BE$ all bisect each other and are thus concurrent.