$AD:DC = BE:EA = 1:2$. Lines $BD$ and $CE$ meet at point $O$. Prove that $\angle AOC = 90^{\circ}$

Question

Points $D$ and $E$ divide sides $AC$ ans $AB$ of an equilateral triengle in the ratio $AD:DC = BE:EA = 1:2$. Lines $BD$ and $CE$ meet at point $O$. Prove that $\angle AOC = 90^{\circ}$

My Work
Take point $F$ in $BC$ so that $CF:FB = 1:2$. The line $AF$ intersects line $BD$ and $CE$ at point $Q$
If we draw lines $FF_1 \parallel BD$ and $FF_2 \parallel CE$ then it is easy to see that $AP:PF = 3:4$ and $AQ:QF = 6:1$. So $AP:PQ:QF = 3:3:1$.
Now how to continue from here. I think I am almost there. But missing something. If I can prove that $PQ = PQ = OQ$ then I am done, as $P$ is the midpoint of $AQ$. But how to do this.
I'd also like to know any over-killing strategy that will simply solve this problem.

Answer 1

Since $BD=CE=AF=k$, we get $OP=OQ=PQ=\frac{3}{7}k$ and since $AP=PQ$,

we get$AP=PQ=PO$ and we are done!

Answer 2

Extend $AO$ to cut $BC$ in $F$ and also let $G$ be the altitude from $C$ to $BC$. Now from Ceva's Theorem we have that $BF:FC = 1:4$. From Menelaus' Theorem on $\triangle BCE$ and $A-O-F$ we have:

$$1 = \frac{BF}{FC} \times \frac{CO}{OE} \times \frac{EA}{BA} \implies \frac{CO}{OE} = 6$$

Now we'll prove that $\angle EDC = 90$. Draw the altitude from $B$ to $AC$ and let it's foot be $X$. It's easy to see that $\frac{AD}{DX} = 2$. So as $\frac{AD}{DX} =2 = \frac{AE}{EB}$ we get that $ED \parallel BX$, so $\angle EDC = 90$.

Finally we have:

$$\frac{AG}{GE} \times \frac{EO}{OC} \times \frac{CD}{DA} = 3 \cdot \frac 16 \cdot 2 = 1$$

Hence $CG, ED, AO$ intersect and they do this at the orthocenter of $\triangle CEA$, as $ED, CG$ are altitudes in the triangle. Hence $AO$ is altitude in $\triangle CEA$ and $\angle AOC = 90$

Answer 3

Triangles say $ABD$ and $BCE$ are congruent because $AB = BC, \,\, AD = \frac{1}{3} AC = \frac{1}{3} AB = BE$ and $\angle \, BAD = \angle \, CBE = 60^{\circ}$. Hence $\angle \, ADO = \angle \, ADC = \angle \, CEB = \angle \, OEB = \theta$. This implies that $\angle \, ADO + \angle \, AEO = \angle \, ADO + 180^{\circ} - \angle OEB = \theta + 180^{\circ} - \theta = 180^{\circ}.$ Therefore quad $ADOE$ is inscribed in a circle and $\angle \, AOE = \angle \, ADE$.

Now, draw a line through point $E$ parallel to $BC$ and denote by $E'$ its intersection point with $AC$. Then triangle $AEE'$ is also equilateral and $AE' = \frac{2}{3} AC$. However, $AD = \frac{1}{3}AC$ so $D$ is the midpoint of $AE'$. Thus $ED$ is the median in the equilateral triangle $AEE'$ from vertex $E$ and is therefore an altitude so $\angle \, ADE = 90^{\circ}$. As already proved $\angle \, AOE = \angle \, ADE = 90^{\circ}$ so $\angle \, AOC = 90^{\circ}$ as well.

Edit. If you want to prove that $PQ= PO =OQ$ you can simply take the center $G$ of triangle $ABC$ (point $G$ is the intersection of the three altitudes, which are the three medians, which are the three angle bisector, which are the three orthogonal bisectors of the edges) and perform a $120^{\circ}$ rotation around it (say counterclockwise). Then triangle $CAF$ is rotated to triangle $ABD$ which in its own turn is rotated to triangle $BCE$. Consequently, segment $AF$ is rotated to $BD$ which is rotated to $CE$. Therefore, since the pair of segments $AF,\, BD$ is rotated to the pair of segments $BD, \, CE$, the intersection point $P = AF \cap BD$ is rotated to intersection point $O = BD \cap CE$. Analogously, you see that point $O$ is rotated to point $Q$. Therefore, the points $P, O, Q$ form an equilateral triangle.

Or, alternatively, since $AF$ is $120^{\circ}$-rotated to $BD$ the intersection angle between the two should be $60^{\circ}$, i.e. $\angle \, FPB = \angle \, QPO = 60^{\circ}$. By the same argument, $\angle \, POQ = 60^{\circ}$ and $\angle \, OQP = 60^{\circ}$. Therefore $OPQ$ is an equilateral triangle.

Answer 4

Let $F$ be the point where the line $AO$ intersects side $BC$.

Applying Ceva's Theorem,

\begin{align*} &\frac{AE}{EB}\cdot\frac{BF}{FC}\cdot\frac{CD}{DA} = 1\\[6pt] \implies\;&\frac{2}{1}\cdot\frac{BF}{FC}\cdot\frac{2}{1} = 1\\[6pt] \implies\;&\frac{BF}{FC}=\frac{1}{4} \end{align*}

Without loss of generality, assume triangle $ABC$ is in $\mathbb{R}^2$, centered at the origin.

Let $a,b,c,d,e,f$ denote the vectors from the origin to the points $A.B,C,D,E,F$, respectively. Then

$$a \cdot a = b \cdot b = c \cdot c$$

$$a \cdot b = b \cdot c = c \cdot a$$

and from the known ratios, we get

\begin{align*} e &= a + (2/3)(b - a)\\[4pt] &= (1/3)a + (2/3)b\\[12pt] f &= b + (1/5)(c - b)\\[4pt] &= (4/5)b + (1/5)c \end{align*}

Then \begin{align*} \overrightarrow{AF}\cdot\overrightarrow{EC} &= (f - a)\cdot(c - e)\\[4pt] &= {\large(}(4/5)b + (1/5)c - a{\large)}\cdot{\large(}c - (1/3)a - (2/3)b{\large)}\\[4pt] &=(1/15)\left({\large(}4b + c - 5a{\large)}\cdot{\large(}3c - a - 2b{\large)}\right)\\[4pt] &=(1/15) \left( {\large(} 8(b\cdot b) - 3(c\cdot c) - 5(a\cdot a) {\large)} + {\large(} 6(a\cdot b) + 10(b\cdot c) - 16(c\cdot a) {\large)} \right)\\[4pt] &=(1/15)(0 + 0)\\[4pt] &=0 \end{align*}

Since $\overrightarrow{AF}\cdot\overrightarrow{EC} = 0$, it follows that lines $AF$ and $EC$ are perpendicular, hence line segments $AO$ and $OC$ are also perpendicular.

Therefore $\angle AOC = 90^{\circ}$.