# area of quadrilateral! only areas given!

Find the $$ar(CEF)+ar(FGB) =\;\;?$$

I am really stuck on this ... spend some hours... did not what to solve and how to proceed? Any suggestions ? Hints also work :)

• This has nothing to do with calculus or functional analysis. Commented May 12, 2019 at 12:18
• Hints? $BD$ goes through $H$ and $DG$ is parallel to $EH$. Commented May 12, 2019 at 12:28
• @EthanBolker: Do you use the given areas to reach those conclusions? They don't appear to be forced by the figure, ignoring the areas. Commented May 12, 2019 at 12:42
• @HenningMakholm On second thought you're probably right. I did question my Hint(?). Commented May 12, 2019 at 13:23

Hint: Try to show that $$S(ABCD) =8 \,S(FEH)$$

Full Solution:

You have $$\vec{FH} = \frac12 \vec{FG} = \frac12 (\vec{BG} - \vec{BF}) = \frac14 (\vec{BA} - \vec{BC})$$ $$\vec{FE} = \vec{FC} + \vec{CE} = \frac12\vec{BC} + \frac12\vec{CD} = \frac12\vec{BD}$$ so $$S(FEH) = \frac12|\vec{FE}\times\vec{FH}| = \frac{1}{16}|\vec{BD}\times(\vec{BA} - \vec{BC})|$$ We also have $$S(ABCD) = S(BCD)+S(BDA) = \frac12|\vec{BC}\times\vec{BD}|+ \frac12|\vec{BD}\times\vec{BA}| = \frac12|\vec{BD}\times(\vec{BA}-\vec{BC})|$$ So $$S(ABCD) =8 \,S(FEH) = 64$$ Therefore $$S(CEF) + S(FGB) = S(ABCD) - S(AGHED) - S(HFE) = 64 - 36 -8 = 20$$

Analytic geometry to the rescue?

We have $$\triangle GHE=\triangle FHE$$, so we can rewrite the data as $$\triangle EFG=16 \qquad \Box ADEG=28$$ and forget about $$H$$ completely.

Next I would select a non-rectangular coordinate system with $$G$$ at $$(0,0)$$, $$E$$ at $$(-1,0)$$ and $$D$$ at $$(0,-1)$$. The coordinates of everything now become simple linear functions of, say, $$x_A$$ and $$y_A$$, and solving $$\frac{\triangle EFG}{\Box ADEG} = \frac{y_F}{1+x_A} = \frac{16}{28}$$ (where the middle fraction rewrites both areas in units of $$\triangle EGC$$ which is half the fundamental parallelogram of the coordinate system) gives us a line that $$A$$ must lie on.

Now we can compute the area $$\triangle CEF + \triangle FGB$$ for an arbitrary point on that line (in the same ad-hoc units). Hopefully it becomes a constant multiple of $$y_F$$ which we know to represent the area $$16$$.

Make the labels as indicated on the figure:

$$\hspace{3cm}$$

Note: \begin{align}&\text{1) JE is the middle line of \Delta ACD and JE||AC.}\\ &\text{2) FG is the middle line of \Delta ABC and FG||AC.}\\ &\text{3) 1) and 2) imply JE||FG} \\ &\text{4) similarly, FE||GJ, hence EFGJ is a parallelogram, whose area is 32 (why?)}\\ &\text{5) S_{ACD}=4S_{DEJ}, S_{ABD}=4S_{AGJ},S_{ABC}=4S_{BFG},S_{BCD}=4S_{CEF}}\\ &\text{6) S_{ABCD}=\frac12(S_{ACD}+S_{ABD}+S_{ABC}+S_{BCD})=2(\underbrace{S_{DEJ}+S_{AGJ}}_{12}+S_{BFG}+S_{CEF})}\\ &\text{7) S_{BFG}+S_{CEF}=S_{ABCD}-44=2(12+S_{BFG}+S_{CEF})-44\Rightarrow S_{BFG}+S_{CEF}=20.}\\ \end{align}

In any quadrangle $$Q$$ the midpoints of the four edges form a parallelogram $$P$$ whose area is $${1\over2}{\rm area}(Q)$$, and the four outer triangles make up the other half of $${\rm area}(Q)$$. In the case at hand $${1\over4}{\rm area}(P)=8$$, so that $$P$$ as well as the four outer triangles together have area $$32$$. The quadrangle $$AGED$$ in the figure has area $$36-8=28$$. This quadrangle consists of the left two outer triangles and one half of $$P$$. These triangles together then have area $$28-{1\over2}{\rm area}(P)=12$$, so that the right two outer triangles together have area $$32-12=20$$.

• Interesting question and interesting answer. +1. We have similar methods, however, mine seems a little too technical and sophisticated. Commented May 17, 2019 at 18:05