Trapezoid geometry problem involving areas 
ABCD is a trapezoid with $AD$ parallel to $BC$. The area of $ABCD$ is $225$. The area of $\triangle BPC$ is $49$. What is the area of $\triangle APD?$ 


I can see that the triangles $BPC$ and $APD$ are congruent, though I do not know how to apply this to a solution. The formula for the area of a trapezium doesn't seem to be applicable here. 
Could someone provide an explained solution which doesn't involve trigonometric functions? 
Thanks for any help you're able to provide - Jazza.
 A: $$ \frac{S_{\triangle ABP}}{S_{\triangle BCP}} = \frac{AP}{PC} = \frac{DP}{BP} = \frac{S_{\triangle DCP}}{S_{\triangle BCP}} := r$$
(Assume this ratio is $r$.) Then we also have
$$ \frac{S_{\triangle ADP}}{S_{\triangle ABP}} = \frac{DP}{BP} = r.$$
Therefore, 
$$S_{\triangle ABP} = 49r $$
$$S_{\triangle DCP} = 49r $$
$$S_{\triangle ADP} = 49r^2 $$
So
$$ 49 + 49r + 49r+ 49r^2 = 225$$ and $$ r= \frac{8}{7}$$
$$S_{\triangle ADP} = 49r^2 = 64$$
A: Let $a=BC$ and $b=AD$ the bases of the trapezoid, $h_1$, $h_2$ and $h=h_1+h_2$ the heights of $PBC$, $PAD$ and $ABCD$. From the similitude of triangles $PBC$ and $PAD$ we have $h_2/h_1=b/a$, so we may write:
$$
225={1\over2}(a+b)h={1\over2}(a+b)(h_1+h_2)
={1\over2}ah_1\left(1+{b\over a}\right)^2=49\left(1+{b\over a}\right)^2.
$$
From that we gain:
$$
1+{b\over a}={15\over7},
\quad\hbox{that is:}\quad
{b\over a}={8\over7},
$$
and finally:
$$
\hbox{area of}\ \triangle APD=
\left({b\over a}\right)^2 \hbox{area of}\ \triangle PBC=64.
$$
A: Revisiting this question, I've got another answer.
In any quadrilateral $ABCD$, where $AC$ and $BD$ intersect at $O$, it is easy to prove that  $|ABO|\cdot |DCO|=|ADO|\cdot |BCO|$ where $|$ denotes area. 
In this question, we see that $|ABP|=|DCP|$ - let this area be $y$, and let $|APD|=x$. Then by the above equality, we have that $y^2=49x\implies y=7\sqrt{x}$. Since $|ABCD|=225, x+14\sqrt{x}+49=225$, and so $(\sqrt{x}+7)^2=225$. Hence $\sqrt{x}=8$ and thus $|APD|=64$.
