Question on inscribed equilateral triangle

Question

Question:

$ABC$ and $ODE$ are equilateral triangle with $BC || DE$. If $O$ is the center of the circle, then find the ratio $AQ:QC$

So, my thought on this is that, since we are not given the length, let $AO = x$

Then, the perpendicular $OZ$ falling on line segment $BC$ is $\frac{2}{3}x$ (I am not sure)

So, using $$\text{Area using Heron's formula} = \text{Normal area}$$

$$\Rightarrow \sqrt{\frac{3x}{2} * (3 * (\frac{3x}{2} - x))} = \frac{1}{2}*(\frac{2x}{3} + x) * x$$

$$\Rightarrow \frac{9x^2}{4} = \frac{2x^2 + 3x^2}{6}$$

$$\Rightarrow \frac{17x^2}{12} = 0$$

which gives $x = 0$ (not possible!)

Can anyone please tell what should be the approach to this problem.

Thank you.

Reference Diagram (courtesy Alex R):

Answer 1

This sketch (made with GeoGebra) should do better than a hand-drawn. All Specs are met (both triangles are equilateral, $BC\parallel DE$ and the circle is in fact a circle.

From this we can use the intercept theorem for $PQ\parallel BC$ and thus $AQ:QC = AR:RS$ But since $AR=OR$ by construction, $RO = OS = AR$ and therefore $$AR:RS = 1:2=AQ:QC$$

Answer 2

Calling $Q$ the intersection point of the lines $(AC)$ and $(OE)$, one can check that $AQO$ is isocele in $Q$, since both the angles $\widehat{Q A O}$ and $\widehat{Q O A}$ are $\pi/6$. From this one can compute the length of its sides depending on the length of its basis (do some trigonometry in one of its half-triangles).

Computing $AO$ from $AC$ is a classical exercise of equilateral triangle properties.

So we get both $AC$ and $AQ$ from $AO$, and their quotient (almost) gives the desired result.