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Triangle $ABC$ in the figure has area $10$ . Points $D,E,$ and $F$, all distinct from $A,B,$ and $C$, are on sides $AB,BC,$ and $CA$ respectively, and $AD=2$ , $DB=3$ . If triangle $ABE$ and quadrilateral $DBEF$ have equal areas,then what is that area? enter image description here

Efforts made: I've tried to add some extra lines to see if i could get something usefull but ,guess, i didnt get anything . It seems like the problem is asking some crazy creative thing to be done,i cant see what.

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  • $\begingroup$ I would begin my seting up the system of equations from your given constraints. You know that the area of the large triangle is 10 so 10=5/2*h, and that the area of ABE can be set equal to the area of DBEF. How many more equations can you introduce to match the number of unknowns? $\endgroup$ – IPoiler Sep 17 '15 at 8:26
  • $\begingroup$ I've tried to decompose the area of the quadrilateral in question but there are so many variables and that quadrilateral is really an ugly one. $\endgroup$ – Nameless Sep 17 '15 at 8:28
  • $\begingroup$ $DF$ and $AE$ intersect at $G$. The equal area requirement can be translated to $\triangle ADG$ and $\triangle EFG$ having equal area. $\endgroup$ – Arthur Sep 17 '15 at 8:29
  • $\begingroup$ Hint: $DE$ is parallel to $???$ $\endgroup$ – achille hui Sep 17 '15 at 8:29
  • $\begingroup$ @ Arthur can you please elaborate a little bit more ? $\endgroup$ – Nameless Sep 17 '15 at 8:47
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$[DBEF]=[ABE]$ is equivalent, by subtracting $[DBE]$ to both sides, to $[DEF]=[DEA]$.
These triangles share the $DE$-side, hence $[DEF]=[DEA]$ implies $DE\parallel AF$, so: $$ \frac{BE}{BC}=\frac{BD}{BA}=\frac{3}{5} $$ and the area of $[BDE]$, consequently, equals $\frac{9}{25}[ABC]=\frac{18}{5}$. Since $[ABE]=\frac{5}{3}[DBE]$, $$ [ABE]=[DBEF]=\color{red}{6} $$ follows.

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  • $\begingroup$ Ho fatto stupidi errori di calcolo se no avrei ottenuto anch'io $6$. Comunque grazie mille per la risposta(p.s: scusa se non c'entra con la domanda,ma hai studiato alla normale di Pisa ?) $\endgroup$ – Nameless Sep 17 '15 at 14:55
  • $\begingroup$ @Jhon: you're welcome. By the way, yes, I studied there, but just the first year of my student career. $\endgroup$ – Jack D'Aurizio Sep 17 '15 at 15:05

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