Question

In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

I have no idea how to do this. I did a problem almost exactly the same and approached it the same way, it did not work.

Answer 1

No diagram, so we don't have any idea about F.

You cannot copy and paste here.

Answer 2

We are told that DE = 2EC, which means that DE/EC = 2/1, and DE = ⅔ DC. Since
AB = DC, it follows that DE = ⅔ AB, and DE/AB = 2/3. Because segments AB and DC are
each perpendicular to segment BC, it follows that segment AB and segment CD (or segment
DE) are parallel. Thus, m∠BAF = m∠DEF, and m∠FDE = m∠ABF because they are pairs of
alternate interior angles. By Angle-Angle Similarity, we have ∆ABF ~ ∆EDF. Notice that segment
BG is an altitude of ∆ABF, and segment CG is the corresponding altitude of ∆EDF. Therefore,
CG/BG = 2/3 and BG = 3/5  BC. Right triangles BGF and BCD are also similar (Angle-Angle
Similarity using the right angles and ∠FBG in each triangle), which means that BC/DC = BG/FG.
Substituting and cross-multiplying yields BC/20 = (3/5  BC)/FG → BC × FG = 20(3/5  BC) → FG = 12.

Answer 3

Because $\overline{AB}$, $\overline{FG}$, and $\overline{DC}$ are all perpendicular to $\overline{BC}$, we have $\overline{AB}\parallel\overline{FG}\parallel\overline{DC}$ Therefore, we have $\angle FAB = \angle FED$ and $\angle EDF = \angle FBA$, which means that $\triangle FAB \sim\triangle FED$. So, we have $FB / FD = AB/DE$. Because $DE/DC = 2/3$ and $AB = DC$, we have $FB/FD = AB/DE = DC/DE = 3/2$. Since $FB/FD = 3/2$, we have $FB/BD = 3/5$. We have $\triangle FBG\sim\triangle DBC$ by AA Similarity, so $FG/DC = FB/BD = 3/5$. Therefore, we have $FG = (3/5)DC = \boxed{12}$.