Consider a trapezoid $ABCD$ with bases $AB$ and $CD$ satisfying $| AB | > | CD |$. Let $M$ be the midpoint of $AB$. Let the point $P$ lie inside $ABCD$ such that $| AD | = | PC |$ and $| BC | = | PD |$. Prove that if $| \angle CMD | = 90^o$, then the quadrilaterals $AMPD$ and $BMPC$ have the same area.
Obviously $\text{area}(APM)=\text{area}(BPM)$. So we need to prove $\text{area}(ADP)=\text{area}(BCP)$. By the segment length conditions, it will be enough to show that $\angle ADP+\angle PCB=180^\circ$. If we let $X=AD\cup BC$, we know $X$ is on the opposite side of $CD$ as $BC$ because $AB>CD$, so we just need to show that the quadrilateral $PCXD$ is cyclic.
Re-define $P$ to be the second intersection of the circle $(DXC)$ and line $XM$. We would like to show $AD=PC$ and $BC=PD$, and this will be enough to imply this point identifies with the original $P$ because $P$ is the only point satisfying the condition on its side of the line $CD$.
For this, let N be the midpoint of CD. Then $X, N, M$ are collinear, and $MN=DN=CN$ by $\angle CMD=90^\circ$. To prove that $BC=DP$ for example, we calculate:
\[\frac{BC}{XC}=\frac{MN}{XN}=\frac{DN}{XN}=\frac{DP}{XC}\]The last equality follows by $X, C, D, P$ concyclic. $\blacksquare$
If we change the word "trapezoid" for "quadrilateral (concave or convex), whose opposite sides $AB$, $CD$ and $BC$, $AD$ don't intersect" the theorem remains true.