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In the figure below, $BD=DE=EC$, $F$ divides $AD$ so that $FA:FD=1:2$ and $G$ divides $AE$ so that $GA:GE=2:1$.

Find ratio of area of triangle $\Delta{AFG}$ to area of $\Delta {ABC}$

enter image description here

My Try:

I noticed that $G$ is centroid of $\Delta{ADC}$ and

$$Ar(ABD)=Ar(ADE)=Ar(AEC)$$

any clue?

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We have that $$|AFG|\stackrel{|AD|=3|AF|}{=}\frac{1}{3}\cdot |AGD|\stackrel{3|AG|=2|AE|}{=}\frac{1}{3}\cdot\frac{2}{3}\cdot |ADE|\stackrel{|BC|=3|DE|}{=}\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot |ABC|.$$

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HINT:

Lemma: Prove that $S_{ABC}=\dfrac{1}{2}AB\times AC\times\sin{BAC}$ and apply similarly for other angles

I will use the lemma above, but I will not prove it here.

Firstly, notice that $S_{ADE}$ has the same altitude as $S_{ABC}$, but $\dfrac{DE}{BC}=\dfrac{1}{3}$, so using the area formula $\dfrac{\text{base}\times\text{height}}{2}$, we will have $\dfrac{S_{ADE}}{S_{ABC}}=\dfrac{1}{3}$ (this step have not used the lemma yet). $\Rightarrow S_{ADE}=\dfrac{1}{3}S_{ABC}$

Secondly, using the lemma above, we can prove that $S_{ADE}=\dfrac{1}{2}AD\times AE\times\sin{DAE}$ and $S_{AFG}=\dfrac{1}{2}AF\times AG\times\sin{DAE}$

$\Rightarrow \dfrac{S_{AFG}}{S_{ADE}}=\dfrac{\frac{1}{2}AF\times AG\times\sin{DAE}}{\frac{1}{2}AD\times AE\times\sin{DAE}}=\dfrac{AF}{AD}\times\dfrac{AG}{AE}$

Hope these hints are enough.

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  • $\begingroup$ Yes understood thanks $\endgroup$ May 16 '18 at 14:09
  • $\begingroup$ Is it possible without Trigonometry $\endgroup$ May 16 '18 at 14:10
  • $\begingroup$ Yes, it is possible: Draw altitudes $FH$ and $DI$ of $\Delta{AFG}$ and $\Delta{ADE}$ respectively. Because of the intercept theorem, $\frac{HF}{AF}=\frac{ID}{AD}$ and we can prove that $S_{AFC}=\frac{AG\times HF}{2}=\frac{1}{2}\times AF\times AG\times \frac{HF}{AF}$ and $S_{ADE}=\frac{AE\times DI}{2}=\frac{1}{2}\times AE\times AD\times\frac{ID}{AD}$ $\endgroup$
    – user061703
    May 16 '18 at 14:33

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