Beowulf181

Ok my instructor helped me along with this problem. In case any of you are interested here is the solution. Thanks for the help!

Oh just so you guys know that angle that you were talking about earlier (angle CDE) the angle is free. the proof is for all values that make that general shape. The only restriants are the parallel lines, perpendiculars, basically the begining instructions on how to construct the object. Thanks.

This is a simpler screenshot of the same basic shape

BTW, help requested is hint on how to proceed, not the solution.

Here is another showing hidden objects like circles used to construct the two.

See attached.

Oh and if anyone has Geometers sketchpad 4 I can email you a picture of the figure.

Ok this is my first time posting on this site, If anyone can help me with this that would be great. Its kinda hard to explain but ill give it a try. There is a Square (ABCD) with a segment extending from the bottom left hand point (B) a bit more then two times the length of the side (BC) of the square. the point ending this segment (H) has a segment extending from that to the upper left hand point of the square (D) and past that point to a the intersection point (E) of that line and a line perpendicular to that extending from the upper right hand corner of the square (A). There is a segment begining at (A) and ending at (E). There is another point (G) that lies on the intersection of segment (HB) and the line extending from point (A) that is perpendicular to segment (EA). the last point (F) lies on the intersection of segment (HE) and the line perpendicular to segment (HE) extending to (G) the final figure should be a quadrilateral with a square and a rectangle inscribed in it. the goal is to prove that the square is equal in area to the rectangle. Once again if you could help that would be great.