Geometry Math Contest Problem

Let the area of triangle $APD$ be $x$ and the area of triangle $DPC$ be $y$ ,

Since $AD=BC$ and $CD=AB$ ,

Area of triangle $APD +20 =$ Area of triangle $DPC +12$ as two pairs of triangles sum to half of the area of parallelogram

$\Rightarrow x+20=y+12 \Rightarrow y=x+8$

Half of the area of parallelogram $= \frac{12+20+x+x+8}{2} = 20+x$

Area of triangle $BDP = 20+8+x-(20+x) = 8$

Let $h$ be the height of parallelogram $ABCD$ , $b$ be the base of parallelogram $ABCD$ , and $x$ be the height of $\triangle APD$ .

Then $A_{\triangle ABD} = \frac{1}{2}bh$ , $A_{\triangle PBC} = \frac{1}{2}bx$ , and $A_{\triangle APD} = \frac{1}{2}b(h - x) = \frac{1}{2}bh - \frac{1}{2}bx = A_{\triangle ABD} - A_{\triangle PBC}$ .

Since $A_{\triangle ABD} = A_{\triangle BDP} + A_{\triangle APB} + A_{\triangle APD}$ and $A_{\triangle APD} = A_{\triangle ABD} - A_{\triangle PBC}$ , $A_{\triangle BDP} = A_{\triangle PBC} - A_{\triangle APB} = 20 - 12 = \boxed{8}$ .