# Having trouble with Geometry problem

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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.

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

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i just tried drawing it but couldn't understand most of it. perhaps you could try drawing the equation in MS paint or another drawing program?

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if you can make a picture of it to send us, then just take a screenshot of it and post that. there is no way we are going to be able to recreate it unless you go into the lengths and angle measurements.

i cant figure out which direction BH is going.

See attached.

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Here is another showing hidden objects like circles used

to construct the two.

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

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This is a simpler screenshot of the same basic shape

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I don't think its drawn right, where did you get the problem?

angle cde should be 45

because, a line that bisects bcd should be equal to y

however more importantly it would make it 45 45 90 and angle cde is not 54 because it does not intersect with point b

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I don't think its drawn right' date=' where did you get the problem?

angle cde should be 45

because, a line that bisects bcd should be equal to y

however more importantly it would make it 45 45 90 and angle cde is not 54 because it does not intersect with point b[/quote']

cde should be less than 45 because "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)"

if it were exactly double then the angle would be 45.

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yeah youre right I just looked it over again I assumed something I shouldn't have

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Typical, it's over before I even look

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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.

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

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