# Find the area of a figure in a triangle

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?

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.

• 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? – IPoiler Sep 17 '15 at 8:26
• 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. – Nameless Sep 17 '15 at 8:28
• $DF$ and $AE$ intersect at $G$. The equal area requirement can be translated to $\triangle ADG$ and $\triangle EFG$ having equal area. – Arthur Sep 17 '15 at 8:29
• Hint: $DE$ is parallel to $???$ – achille hui Sep 17 '15 at 8:29
• @ Arthur can you please elaborate a little bit more ? – Nameless Sep 17 '15 at 8:47

$[DBEF]=[ABE]$ 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.
• 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 ?) – Nameless Sep 17 '15 at 14:55