WLOG $A = (0,0),B(1,0) , C(m+1,n) ,D(m,n)$ Lol bash quite clean here since $P \in AB$ implies $P = (k,0)$ gradient BC is $\frac{n}{m} = \nabla{BC}$ the line that is pass through p and parallel gonna have same gradient so the line has eqn $y-0 = (x-k)*\frac{n}{m}$ Line Equation AC could be computed easily to be $y = \frac{xn}{m+1}$ Equalizing the y we get the coordinate Q at $(k(m+1),kn )$ Use Shoelace DAQ we get $[DAQ] = \frac{|kn|}{2}$ $[PAQ]$ is easily $\frac{k*kn}{2} = \frac{Base *height}{2}$ same wise with $[BCD]$ which is $\frac{n}{2}$ $\frac{k^2n^2}{4} = \frac{k^2n}{2}*\frac{n}{2}$ as required

It can be also solved just by moving the areas. See that $[DAQ] = [DAP] = \frac{1}{2}AP \cdot H$, $[PAQ] = \frac{1}{2} AP \cdot h$ and $[BCD] = [ABC] = \frac{1}{2} AB \cdot H$, where $h$ and $H$ are the distances from $C$ and $Q$ to $AB$. Then we can see that we need to prove $\frac{1}{4}AP^2 H^2 = \frac{1}{4}AP \cdot AB \cdot H \cdot h$, which is $ \frac{AP}{h} = \frac{AB}{H}$, but this is just Thales' theorem $(BC \parallel PQ)$.