# Prove that quadrilateral $ADOE$ is cyclic

Let there be a triangle $ABC$ such that $\angle BAC = 60$. Points $D$ and $E$ bisect sides $AB$ and $AC$, respectively. If $O$ is a point in the interior of $\triangle ABC$ such that $\angle AOB = \angle BOC = \angle COA = 120$, prove that quadrilateral $ADOE$ is cyclic.

I drew this in GeoGebra and it definitely seems to be true. I am trying to figure out a way to relate the midpoints of the sides and so I drew the medial line segment to them but am unsure how this helps. It would suffice to show that $\angle{DOE} = 120^{\circ}$.

First of all, observe that, since there is no angle in $\triangle ABC$ that is greater than $120^o$, $~ O$ is the Fermat point of the triangle.

Notice that $\angle MOL = 120^o$ so it suffices to prove that $\angle DOE = \angle MOL$, or that $\triangle MDO$ is similar to $\triangle LEO$.

Now use that $M,A,L$ are collinear and with the help of some similar triangles prove that $\angle DMO = \angle OLE$.

You can finish the proof by using ratios and similar triangle to prove that $\frac{MD}{LE} = \frac{MO}{LO}$.

Enjoy!

• How do you know that $O$ is the Fermat point? Commented Jan 24, 2016 at 21:15
• It's straightforward from the fact that $\angle BOC=\angle AOC=\angle AOB = 120^o$. Only the Fermat point has this property and from its construction it follows that it is unique.
– Greg
Commented Jan 24, 2016 at 21:20
• Why does $\angle{DMO} = \angle{OLE}$ Commented Jan 24, 2016 at 23:42
• $\triangle MBO \sim \triangle LBM \implies \angle BMO = \angle MLO \implies \ldots$
– Greg
Commented Jan 25, 2016 at 15:24
• I'll post my proof as an answer. Commented Jan 25, 2016 at 21:10

Since $\triangle{ABC}$ is a triangle with angles less than $120^{\circ}$, $O$ is the Fermat point of $\triangle{ABC}$. In the diagram below, $\angle{MOL}$ is $120^{\circ}$ since $\angle{MOB} = 60^{\circ}$. It then suffices to show that $\angle{DOE} = \angle{MOL}$, or that $\triangle{MDO} \sim \triangle{LEO}$. Now since $M,A,$ and $L$ are collinear, see that $\triangle{MBO} \sim \triangle{LBM}$. As a result, $\angle{BMO} = \angle{MLO} \iff\angle{DMO} = \angle{OLE}$. Finally, see that since $\triangle{AMO} \sim \triangle{CLO}$ (this is true since $\angle{AOM} = \angle{LOC} =60^{\circ}$ and $\angle{ACO} = \angle{OAB}$ and $\angle{CLO} = \angle{OMA}$ since $\angle{BMO} = \angle{MLO} \iff \angle{CLO} = \angle{OMA}$), then $\dfrac{MD}{LE} = \dfrac{MO}{LO}$, we have that $\triangle{MDO} \sim \triangle{LEO}$. Thus, since $\angle{MOL} = 120^{\circ}$ and $\angle{DOM} = \angle{LOE}$, $\angle{DOE} = 120^{\circ}$ implying that $ADOE$ is cyclic. $\blacksquare$