Find the area of quadrilateral in triangle Let $ABC$ be a triangle with area $12$, $D$ is a point in $AB$ and $E$ a point in $AC$ such that, if $P$ is the intersection of $DC$ and $BE$, the triangle $BPD$, the triangle $CPE$ and the quadrilateral $ADPE$ have the same area. Find the area of $ADPE$. 
First I have seen that the segment $DE$ is parallel to $BC$ and that the triangles $ADE$ and $ABC$ are similar in the same ratio that the triangles $DEP$ and $CBP$ are similar, but I don't know how to finish with this information.
 A: Let 
$$x=[\triangle{BPD}]=[\triangle{CPE}]=[\square ADPE],\quad y=[\triangle{PBC}]$$
Then, we have
$$3x+y=12\tag1$$
Now, 
$$[\triangle{CDA}]:[\triangle{ABC}]=2x:12\tag2$$
Since $\triangle{CDA}$ and $\triangle{ABC}$ have a common height, we get
$$[\triangle{CDA}]:[\triangle{ABC}]=AD:AB\tag3$$
You've already noticed that $DE$ is parallel to $BC$, so $\triangle{ADE}$ and $\triangle{ABC}$ are similar to have
$$AD:AB=DE:BC\tag4$$
Since $\triangle{PDE}$ and $\triangle{PCB}$ are similar,
$$DE:BC=PD:PC\tag5$$
Since $\triangle{BPD}$ and $\triangle{PBC}$ have a common height, we get
$$PD:PC=[\triangle{BPD}]:[\triangle{PBC}]=x:y\tag6$$
From $(2)(3)(4)(5)(6)$, we have
$$2x:12=x:y\tag7$$
It follows from $(1)(7)$ that $$[\square ADPE]=x=2$$
A: Not really an answer : I couldn't comment any other way, and thought that an additional point deserved to be made.  $\triangle BPD$ has the same area as $\triangle CPE.$ This implies that $\triangle BCD$ has the same area as $\triangle BCE.$  Area of $\triangle$ = (1/2) base * height, and the latter two triangles have the same base, BC.  Therefore, the latter two triangles have the same "height".  That is, the distance from E to BC must equal the distance from D to BC.  Therefore, DE, the altitude from D to BC, the altitude from E to BC, and the portion of BC between the two altitudes forms a rectangle.  Therefore, DE must be parallel to BC.
