Find the area of quadrilateral in triangle

Question

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.

Answer 1

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 $$[\square ADPE]=x=2$$

Answer 2

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.