trial solution

O is the middle point of AB. If cutting along PQ, the lower shape is small than the upper shape, and the difference is related to S0=area(OPC). An area equivalent to S0 needs to be removed from upper shape and added to lower shape. That is, we can push CQ to DQ for adjustment, and deliberately let S1=area(DCQ)=S0. That is, PC/CQ=DC/CO. It is possible to find D by using ruler and compasses.