Area of a triangle from some of its parts

This was not likely to have been an SAT practice problem, though it is a typical contest problem.

$AE:EC = x:y$ (since those two triangles have the same altitude to $AC$, the ratios of their areas is the ratios of their bases with respect to that altitude) and $AP:PD = (x+y):z$ (same idea as $AE:EC$). Knowing these two ratios, apply the technique of mass points, putting masses $zy$ at $A, zx$ at $C$ (gives the ratio $x:y$ for $AE:EC$), and $y(x+y)$ at $D$ (gives $(x+y):z$ for $AP:PD$). This results in a mass of $y(x+y)-zx$ at $B$, so the ratio $BD:DC = zx:(y(x+y)-zx).$ This must also be the ratio of the areas of △ABD to △ADC (common altitude again), so (area of △ABD):$(x+y+z) = zx:(y(x+y)-zx)$. Solving from there is a matter of bashing out the algebra.

Method 1: We can perform an affine transformation to make ADC any triangle we wish, while preserving colinearality and the ratios of areas. In particular, this allows us to assign exact coordinates to A, D and C. We can then use the ratio of x and y to find E and the ratio of z and x+y+z to find P. We can then use these coordinates to find B. Once we know B, we know the ratio of BC to DC and hence the ratio of the area of the whole triangle to the known areas.

Method 2: Let Q be the area of the whole triangle

AE:EC=x:y=|AEB|:|BEC|
|APB|=xQ/(x+y)-x
|BPD|=yQ/(x+y)-y-z
Now AP:PD=x+y:z
z(xQ/(x+y)-x)=(x+y)(yQ/(x+y)-y-z)
Q(zx/(x+y)-y)=xz-xy-y^2-xz-yz=-xy-y^2-xz
Q(zx-xy-y^2)/(x+y)=(-xy-y^2-xz)
Q=(-xy-y^2-xz)*(x+y)/(zx-xy-y^2)