Find the area of BGHF

Question

This problem is taken from the Chinese WMTC 2019 Junior Division.

If ABCD is a square of area 40, and $BE=\frac{1}{3}AB, BF=\frac{2}{5}BC$, find the area of BGHF.

I tried using the idea of area ratios (same-base, same-height), but was stuck.

Using Geogebra, I found that EG = GH and EH = HC. Furthermore, area is 3.

What idea or concept am I missing to get to the fact the EG = GH?

Answer 1

The following is your diagram with the horizontal line $JK$ through $G$, plus the vertical lines $GL$ and $HM$, added to it:

Define $s = \sqrt{40}$ to be the square's side lengths. Next, since $\measuredangle GEB = \measuredangle GCD$, $\measuredangle GBE = \measuredangle GDC$ and $\measuredangle EGB = \measuredangle CGD$, then $\triangle EGB \sim \triangle CGD$, with a similarity ratio of $\lvert EB\rvert$ to $\lvert DC\rvert$, i.e., $1$ to $3$. Thus, their altitudes, as well as $\lvert EG\rvert$ compared to $\lvert GC\rvert$, have the same similarity ratio, so

$$\lvert GJ\rvert = \frac{s}{4}, \;\; \lvert GK\rvert = \lvert LC\rvert = \frac{3s}{4}, \;\; \lvert GC\rvert = 3\lvert EG\rvert \tag{1}\label{eq1A}$$

From $\triangle EBC \sim \triangle GLC$, we get

$$\frac{\lvert EB\rvert}{\lvert BC\rvert} = \frac{\lvert GL\rvert}{\lvert LC\rvert} \;\;\to\;\; \frac{1}{3} = \frac{\lvert GL\rvert}{\frac{3s}{4}} \;\;\to\;\; \lvert GL\rvert = \frac{s}{4} \tag{2}\label{eq2A}$$

Letting $x = \lvert HM\rvert$ then, since $\triangle HMC \sim \triangle EBC$, we have $\lvert MC\rvert = 3x$. Also, $\lvert FM\rvert = \lvert FC \rvert - \lvert MC\rvert = \frac{3s}{5} - 3x$. Since $\triangle HMF \sim \triangle DCF$, then

$$\frac{\lvert FM\rvert}{\lvert HM\rvert} = \frac{\lvert FC\rvert}{\lvert DC\rvert} \;\;\to\;\; \frac{\frac{3s}{5} - 3x}{x} = \frac{3}{5} \;\;\to\;\; \frac{s}{5} - x = \frac{x}{5} \;\;\to\;\; x = \lvert HM\rvert = \frac{s}{6} \tag{3}\label{eq3A}$$

This shows $\lvert EB\rvert = 2\lvert HM\rvert$. Since $\triangle EBC \sim \triangle HMC$, then $\lvert EC\rvert = 2\lvert HC\rvert \;\to\; \lvert EH\rvert = \lvert HC\rvert$. This, along with the last part of \eqref{eq1A}, gives that $\lvert EG\rvert = \lvert GH\rvert$, confirming your Geogebra results.

Finally, using $A$ to be the area function, then since $A(BGHF) = A(BGC) - A(FHC)$, we get using the triangle altitudes in \eqref{eq2A} and \eqref{eq3A} that

$$A(BGHF) = \frac{1}{2}(s)\left(\frac{s}{4}\right) - \frac{1}{2}\left(\frac{3s}{5}\right)\left(\frac{s}{6}\right) = s^2\left(\frac{1}{8} - \frac{1}{20}\right) = 40\left(\frac{5 - 2}{40}\right) = 3 \tag{4}\label{eq4A}$$

Answer 2

L is the center of the square. If $FH=HB$, then and LH is parallel with AC and we have:

$LH=\frac12 FC$

Also:

$AF=\frac 12 FC$

$\Rightarrow LH=AF$

So triangles GAF and GLH are congruent, therefore $FG=GH$.

Answer 3

Our attempt here is to apply Menelaus's theorem to several triangles present in the given configuration to solve this problem. In view of this, two auxiliary lines $AC$ and $FG$ were added to the configuration. The point of intersection of the two diagonals $AC$ and $BD$ of the square $ABCD$ is marked as $O$. Since $O$ is the midpoint of the diagonal $BD$, we have, $$OD : DB = 1 : 2. \tag{1}$$

Now, apply the mentioned theorem to $\triangle ABO$ and the transversal $CE$ to obtain, $$\dfrac{AE}{EB}\times\dfrac{BG}{GO}\times\dfrac{OC}{CA}=\dfrac{2}{1}\times\dfrac{BG}{GO}\times\dfrac{1}{2}=1\qquad\rightarrow\qquad BG : GO = 1 : 1. \tag{2}$$

Equation (2) implies that $G$ is the midpoint of the segment $OB$. From equation (1) and (2), we can deduce that,, $$DG : DB = 3 : 4. \tag{3}$$

Next, we apply the same theorem to $\triangle BCG$ and the transversal $DF$ to obtain, $$\dfrac{BF}{FC}\times\dfrac{CH}{HG}\times\dfrac{GD}{DB}=\dfrac{2}{3}\times\dfrac{CH}{HG}\times\dfrac{3}{4}=1\qquad\rightarrow\qquad CH : HG = 2 : 1.\tag{4}$$

Since the two diagonals of a square divide it into four equal parts, we have, $$\triangle BOC = \dfrac{1}{4} ABCD = \dfrac{40}{4} = 10.$$

According to (2), $CG$ divides $\triangle BOC$ into two equal parts. Therefore, we shall write, $$\triangle BCG = \dfrac{1}{2} \triangle BOC = \dfrac{10}{2} = 5.$$

Since $BF : FC = 2 : 3$, $FG$ divides $\triangle BCG$ to produce $\triangle BFG$ and $\triangle FCG$ such that, $$\triangle BFG = \dfrac{2}{5} \triangle BCG = \dfrac{2\times 5}{5} = 2 \qquad\text{and} \tag{5}$$ $$\triangle FCG = \triangle BCG - \triangle BFG = 5 – 2 = 3.\tag{6}\quad $$

According to (4) and (6), $FH$ divides $\triangle FCG$ to produce $\triangle FHG$ such that, $$\triangle FHG = \dfrac{1}{3} \triangle FCG = \dfrac{3}{3} = 1. \tag{7}$$

Finally, using (5) and (7), we can state, $$DFHG = \triangle BFG + \triangle FHG = 2 + 1 = 3. $$