# Prove that three segments in a convex hexagon intersect at one point

Let $ABCDEF$ be convex hexagon such that $\angle ABC=\angle BCD=\angle DEF=\angle EFA$ and $|AB|+|DE| =|AF|+|CD|$.

Prove that line $l_{AD}$ and perpendicular bisectors of sides $\overline{BC}$ and $\overline{EF}$ intersect at some point.

This is the diagram: As you can see, I prolonged segments $\overline{CD}$ and $\overline{AB}$ to intersect at $X$ and, similarly, $\overline{DE}$ and $\overline{AF}$ to intersect at $Y$. Then, as $\angle ABC=\angle BCD$, we have that triangle $\triangle BXC$ is equilaterial and the perpendicular bisector of the side $\overline{BC}$ is in fact the bisector of the angle $\angle BXC$, and, similarly, perpendicular bisector of $\overline{FE}$ bisects $\angle EYF$.

But what now? Any hints? Let $P$ and $Q$ be points on $\overline{AB}$ and $\overline{AF}$, respectively, such that $|BP| = |CD|$ and $|FQ|=|ED|$. By the segment-length condition, we have $$|AP| = |AB| -\color{blue}{|PB|} = |AB|- \color{blue}{|CD|} = |AF| - \color{red}{|DE|} = |AF| -\color{red}{|FQ|} = |AQ| \tag{1}$$

Moreover, $\square PBCD$ and $\square QFED$ are isosceles trapezoids with congruent "upper base" angles, and therefore also congruent "lower base" angles, so that $\angle APD \cong \angle AQD$. Importantly, these last two angles are not acute. (Why?)

By SSnaA (Side-Side-non-acute-Angle), $\triangle APD \cong \triangle AQD$. We deduce, then, that the perpendicular bisectors in question concur along $\overleftrightarrow{AD}$ simply because the trapezoids are mutual reflections across that line. $\square$

Let me write another solution. Using your picture, $XC=XB$ and $EY=FY$. It follows that $$AX+DY=BX+AB+ED+EY=CX+AF+CD+FY=AY+DX,$$ therefore quadrilateral $AXDY$ is circumscribed. Let $I$ be the center of the circle inscribed in $AXDY$.

Note that the perpendicular bisectors of $BC$ and $EF$ are angle bisectors of $BXC$ and $EYF$. This is because triangles $BXC$ and $EYF$ are isosceles. Thus we only need to prove that $I$ lies on $AD$.

Let $\alpha$ be the common value of angles $ABC, BCD, DEF, EFA$. The internal angle $AID$ in the pentagon $AIDEF$ is equal to $$540^\circ - \angle IDE - \angle DEF - \angle EFA - \angle FAI = 540^\circ - \frac 12 \angle CDE - 2\alpha - \frac 12 \angle FAB.$$ Similarly , the internal angle $DIA$ in the pentagon $DIABC$ is equal to $$540^\circ - \angle IAB - \angle ABC - \angle BCD - \angle CDI = 540^\circ - \frac 12 \angle FAB - 2 \alpha - \frac 12 \angle CDE.$$ It follows that these two angles are equal. On the other hand, they add up to $360^\circ$, so $\angle AID = 180^\circ$. Therefore $I$ lies on $AD$.
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