A question related to triangles , areas , ratio of areas of triangles.

Question

I know the title is confusing but that is because of 150-character limit, if anyone of you can improve it , please do.

Consider $\triangle ABC.$ Choose a point $D$ on segment $BC$ such that $BD/DC=1/2$. Choose a point $E$ on segment $AC$ such that $AE/EC =2/3$ .
Let segments $AD$ and $BE$ intersect at point $P$.
If area of $\triangle PBD = 5$ sq. units, then find the area of quadrilateral $PDCE$.

Here is a sketch that I drew:

My attempt:

Let $A(\triangle APB)=x$ , $A(\triangle APE)=y$ , $A$( quadrilateral $PDCE$)=$z$
Then,
$(x+5)/(y+z)=1/2$ , $(x+y)/(5+z)=2/3$.
So,
$(2x+10)=(y+z)$
And,
$(3x+3y)=(10+2z)$ Hence,
$y = x/5+6, z = ((9 x)/5)+4$
But so what ? I want actually the numerical value of $z$ . What should I do now? Any hints are apreciated. (This is not class-homework , I'm solving sample questions for a competitive exam )

Answer 1

$z = A(\triangle CDP) + A(\triangle CPE) = 2A(\triangle BDP) + \tfrac32 A(\triangle APE) = 10 + \tfrac32 y$

Answer 2

Apply menelaus. BTW, What's your name?

Answer 3

Up to an affine scaling we may assume $$C=(0,0), \quad E=(3,0),\quad A=(5,0),\quad D=(0,2),\quad B=(0,3)\ .$$ This leads to $P=\bigl({5\over3},{4\over3}\bigr)$, so that $$A(\triangle_{BCE})={9\over2},\quad A(\triangle_{BDP})={5\over6}\ .$$ From this we obtain $$A(\square_{\,CEPD})={9\over2}-{5\over6}={22\over 6}={22\over 5}A(\triangle_{BDP})\ .$$ The answer to the question therefore is $22$ square units.