On the products of lengths of adjacent sides in a cyclic quadrilateral

Question

I'm trying to fill in the details of a proof for which I only have a figure, as shown below:

...and a claim that refers to this figure¹:

$$ \overline{EG} : \overline{GF} = ΔEDC : ΔFCD = \overline{ED}·\overline{EC} : \overline{FC}·\overline{FD} $$

I cannot justify the second equality above.

Also, I would like to know if the claim is true for any cyclic quadrilateral $ECFD$ such that the angles at the opposing vertices $C$ and $D$ are right angles (as opposed to requiring some of the other details given in the figure).

---

Here's some additional information that may be useful. First, the segment $AB$ is the diameter of the semicircle that also passes over $D$ and $C$. Next, it is not difficult to show that triangles ACE and BDE are similar right triangles. From this, together with the remaining lengths shown in the figure, one can deduce that (a) $\overline{EC} = 4$, and, therefore, (b) the acute angles of these similar right triangles are $30^\circ$ and $60^\circ$. (I don't know, however, how any of this relates to the claim above, if at all.)

---

¹ The notations $ΔEDC$ and $ΔFCD$ stand for the areas of the triangles $EDC$ and $FCD$, respectively.

Answer 1

I cannot justify the second equality above.

Using the trigonometric formula for area of $ABC$ that $\Delta ABC=\frac{1}{2}bc\sin A$ $$\frac{\Delta EDC}{\Delta FDC}=\frac{\frac{1}{2}ED\cdot EC\sin E}{\frac{1}{2}FD\cdot FC\sin F}=\frac{ED\cdot EC}{FD\cdot FC}\quad \Leftrightarrow \sin E=\sin F$$ This means either $\angle E=\angle F$ or $\angle E+\angle F=180^\circ$ as in a cyclic quadrilateral. Clearly the claim about ratios will hold for any cyclic quadrilateral.