SOLUTION: Hi I am a geometry student in need of dire help. I am really confused with a couple of proofs here is are a few examples. If you could get back to me with a 2 column proof for them

Question 250767: Hi I am a geometry student in need of dire help. I am really confused with a couple of proofs here is are a few examples. If you could get back to me with a 2 column proof for them I would be extremely grateful.
1) ABCD is a parallelogram with angle A acute. A line from B is perpendicular to AD at E, and a line from D is perpendicular to BC at F. AF cuts BE at H, and CE cuts DF at K. Prove EHFK is a parallelogram

2) E is any point on side DC of parallelogram ABCD. DC is extended through C to F, so CF=DE. AE is extended through E to G so EG=AE. Prove EGFB is a parallelogram

Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
I'll try to help with the first problem. Maybe you can resubmit the second if you still need help with it.
To prove that EHFK is a parallelogram we need to show that EH is parallel to KF and also that HF is parallel to EK.
EH is parallel to KF because they are segments on EB and DF respectively and those two lines are parallel because they are both perpendicular to the same line ED.
Remember when a transversal cuts two lines, if corresponding angles are equal in size, or if alternate angles are equal in size, then two lines are parallel.
We still need to show lines HF and EK are parallel.
A big step to showing this is to show that triangles AFD and EBC are congruent.
We can use SAS to show this.
BC and AD are equal because they are opposite sides of the original parallelogram.
BE and FD are equal because they are both altitudes of the original parallelogram
The angle between these two sides in both cases is a right angle so we have SAS.
We only need to show now that angle AFD is equal to EKD and since they intersect the line DF in the same angle HF and EK must be parallel.

Angle BEC and AFD are corresponding angles of the congruent triangles so they are equal.
Angle EKD and BEC are alternate angles of the tranversal EC of the parallel lines EB and FD so angle BEC = angle EKD.
Since angle AFD corresponds (SAS above) to angle BEC they are equal. So we have angle AFD = BEC = EKD (transitive law).
So we have line HF parallel to EK.
EHFK is a paralleogram.

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