# Angle bisector in a trapezoid - surface area ratio

In trapezoid ABCD (AB || CD) the angle bisector of angle ABC is perpendicular to segment AB and intersects it in point P. Point P divides the side AD in ratio 2:1. Find the ratio of the surface areas of the triangle and the quadrilateral.

$h = \frac{2}{3}H$
$S_{triangle}= \frac{1}{3}|AB|H$
$S_{quadrilateral} = S_{trapezoid} -S_{triangle}$
$S_{trapezoid}=\frac{1}{2}(|AB|+|CD|)H$
Now I am stuck. I need to find CD in terms of H and AB, but I simply don't know how to do this. How should I proceed to solve this in the easiest possible way?

Let $E$ be the point where the lines $AD$ and $BC$ intersect.

Let $k = [ABE]$.

In $\Delta ABE$, segment $BP$ is both an angle bisector and an altitude, hence it's also a median.

It follows that $[APB] = [EPB] = \frac{1}{2}k$.

Since $CD\,||\,AB$, it follows that $\Delta DCE$ is similar to $\Delta ABE$.

Since $AP = EP$ and $\displaystyle{{\small{\frac{DP}{AP} = \frac{1}{2}}}}$, it follows that $\displaystyle{{\small{\frac{DE}{AE} = \frac{1}{4}}}}$, hence $[DCE] = \frac{1}{16}k$.

Then $[BCDP] = [EPB] - [DCE] = \frac{1}{2}k - \frac{1}{16}k = \frac{7}{16}k$.

Therefore $\, \displaystyle{ \frac{[APB]}{[BCDP]} = \frac {\left(\frac{1}{2}k\right)} {\left(\frac{7}{16}k\right)} = \boxed{{\small{\frac{\,8\,}{\,7\,}}}} }$