It seems my posts to 'geometry.puzzles' don't get out.I send you in private what I've tried to post there.
1) the answer by Steve is based upon the assumption that the required answer is independant on E. However this can be proved instead :
For any DEFG the triangle CDE is half the area of ABCD (base = CD and altitude = AD)and is also half of area of DEFG (base = DE, altitude = EF) Q.E.D.
2) For your requirement of a trigonometric answer : let t = angle CDE then DE = AD/cos(pi/2 - t) = AD/sin(t) and with H the projection of C on line DE, FE = CH = CD*sin(t) therefore area of DEFG = DE*EF = (AD/sin(t)) * (CD*sin(t)) = AD*CD = S1
3) A proof by disection, being a visual proof, requires a drawing... see attachment. Self explanatory.
Best Regards.Philippe Chevanne (Philippe 92)
______________________From: Avni Pllana <email@example.com>Newsgroups: geometry.puzzlesSubject: Rectangle Area> Let ABCD be an arbitrary rectangle. Let DEFG be a rectangle > such that E is an arbitrary point on the line through AB, and> FG passes through C. If S1 is the area of ABCD, determine the> area S2 of DEFG.