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I have been asked this question from a junior and could not solve the question in a simple way. I am asking help on this platform.

For a triangle $ABC$, Points $D, E$ on $AB$, where $AD:DE:EB=2:2:1$. Point $H$ on $AC$, where $AH:HC=5:3$. Point $G$ on $BC$, where $BG:GC=1:1$. Point $F$ is the intersection between $DG$ and $EH$.

Given area $ADFH$ = 100 $cm^2$, find area $BEFG$.

The difficulty is I don't see any line parallel, so I can't find any similar triangle.

What I have derived from triangle area formula is area $AEH$: area $ABC$ = $1:2$ and area $BDG$: area $ABC$ = $3:10$. But to get area $DEF$'s ratio, I spend quite a long time finding the ratio $EF:FH$.

This should not be such a hard question, so I want to know if there is an easier way of finding the answer.

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Are vectors allowed? – Michael Hoppe Mar 20 '14 at 6:39
I've found $EF:FH=16:21$. – Michael Hoppe Mar 20 '14 at 7:32
@MichaelHoppe Like to share how to get that ratio? – Mick Mar 21 '14 at 9:08
@Mick Of course, have a look at my “answer”, please. – Michael Hoppe Mar 21 '14 at 11:22
@MichaelHoppe With your help, I am able to complete my proof in geometry. See below. – Mick Mar 24 '14 at 19:27

This is too long for a comment. Let $\overrightarrow{AB}=a$ and $\overrightarrow{AC}=b$. Then $$0=\overrightarrow{AH}+ \overrightarrow{HF}+\overrightarrow{FD}+\overrightarrow{DA}.$$ Now substitute


$\overrightarrow{HF}= x\overrightarrow{HE} =x(-\frac{5}{8}b+\frac{4}{5}a)$,

$\overrightarrow{FD}=y\overrightarrow{GD}=y\bigl(\frac{1}{2}(a-b)-\frac{3}{5}a\bigr)$ and


in that equation to get $$\left(\frac{4}{5}x-\frac{1}{10}y-\frac{2}{5}\right)a + \left(\frac{5}{8}-\frac{5}{8}x-\frac{1}{2}y\right)b=0.$$ From here $x=21/37$.

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The following ratios are assumed:-

(1) area of ⊿AEH : area of ⊿ABC = [⊿AEH] : [⊿ABC] = 1 : 2 {User136705’s effort}

(2) [⊿BGD] : [⊿ABC] = 3 : 10 {User136705’s effort}

(3) EF : FH = 16 : 21 {Michael Hoppe’s effort}

Constructions: Produce BA to X such that XA = EB. Join CX. Join HD. enter image description here

Let [⊿CXA] = 1 (square) unit. Then, by same altitude,

(0.1) [⊿CXB] = 6 units

(0.2) [⊿CAB] = 5 units

BG = GC and BD = … = AD implies

(1.1) DG : CX = 1 : 2 …..{Mid-point theorem}

(1.2) [⊿BGD] = (1/4)[⊿CAB] = … = 3/2 units ….. {ratio of areas of similar objects}

Let $[⊿HDF] = t$ $cm^2$. Then, $[⊿AHD] = 100 – t$ $cm^2$

From (3), $[⊿FDE] = (\frac {16} {21})t$ $cm^2$

∵ $[⊿AHD] = [⊿EHD]$……….{same altitude}

∴ $100 – t = t + (\frac {16} {21})t$

∴ $t = … = 2100/58$ $cm^2$

From (1), $[100 + (\frac {16} {21})(\frac {2100} {58})] cm^2 = (1/2)5$ units

∴ $1$ unit $= (2/5)[100 + \frac {1600} {58}]$ $cm^2$

∴ $[$ quad $EBGF] = [⊿GDB] – [⊿FDE]$

$=(3/2)×(2/5)[100 + \frac {1600}{58}] – (\frac {16}{21})(\frac {2100}{58})$ $cm^2$

$= ... = 48.96551724$ $cm^2$

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You're right, but what have you essentially done besides filling the gaps in my answer? Btw, noticing that area of $DEF=2/37\det(a,b)$ the result is achieved almost immedeately. – Michael Hoppe Mar 24 '14 at 19:41
@MichaelHoppe It is not filling the gaps. It is an alternate (in pure geometric) method of solving the problem. The question was raised by a junior who probably has no training in such advanced (vectoric) approach yet. – Mick Mar 25 '14 at 4:10
You don't need all these constructions: from $EF:EH=16:37$ you get at once $area_{DEF}=(8/37)area_{AEH}=(4/37)area_{ABC}$. – Aretino yesterday
@Aretino Yes. Just want to include the OP's finding. – Mick yesterday

Area of $EBGF=(\det(\overrightarrow{BG},\overrightarrow{BE})+\det(\overrightarrow{EG}, \overrightarrow{EF}))/2$, I'll get for that area $\frac{71}{740}\det(a,b)$. Similar area of $ADFH=\frac{59}{296}\det(a,b)$. If that area is supposed to be $100$, area of $EBGF=\frac{2840}{59}\approx 48.14$.

Edit: area of $ADFH=\frac{29}{148}\det(a,b)$, I was off by $1$, so the final result is $1420/29\approx48.97$.

You may achieve easier by noting that the area of $DEF=\frac{2}{37}\det(a,b)$.

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