Two equilateral triangles rest upon the same horizontal line, as shown in the linked figure below.
The red triangle is tilted so that one corner touches a side of the black triangle at the midpoint of the side. A perpendicular line is drawn from the midpoint of the right side of the red triangle through the point where the red triangle touches the black triangle, until it reaches the left side of the black triangle. The line is divided into a yellow segment and a blue segment at the point where the two triangles touch. What is the ratio between the yellow and blue lines? (Hint: It should be the golden ratio ~1.681....)
Since there's a trigonometry tag, here's such an approach, which yields Blue's answer:
From the OP's picture, I assume that the two triangles are congruent. Then, the sine law implies that in $\triangle BED$: $$\sin \frac{2\pi}{3}:\sin \beta = DE:DB = 2.$$ Thus $$\sin \beta=\frac{\sqrt3}{4},\cos \beta=\frac{\sqrt{13}}4.\tag{1}$$
On the other hand, $$\begin{aligned}\gamma &= \pi - (\angle ADB +\alpha +\angle DEG) \\ &=\pi -\left(\frac\pi2+\alpha+\frac\pi6\right)\\ &= \frac\pi3-\alpha\\ &=\beta. \end{aligned}$$ Thus, $\triangle ADH$ and $\triangle AED$ are similar, and we get $$AD:HD = AE:DE$$ Since $AD=DG$ and $AB=DE$, the ratio we are looking for is $$\frac{AE}{DE} = 1+ \frac{BE}{DE}=1+\frac{\sin\alpha}{\sin 2\pi/3}.$$ The answer follows from (1) and the fact that $$\sin \alpha = \sin \left(\frac\pi3-\beta\right)=\sin \frac\pi3\cos\beta-\cos\frac\pi3\sin\beta.$$