document.write( "Question 837038: I'm working on a proof, and I'm having a lot of trouble. I have a parallelogram, ABCD, there is a line from D to B. There is a quadrilateral on the inside, with points on A and C, and the other two point (E and F), on line DB. Given is that DE=BF, and I have to prove that AFCE is a parallelogram. Can anyone help me with this? Thank you. \n" ); document.write( "

Algebra.Com's Answer #504451 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
Hint: Show that AF = CE by showing that triangles AFB and CED are congruent. Similarly, show that AE = FC. That should be sufficient to prove that AFCE is a parallelogram, but you can also show that the opposite interior angles are equal if you aren't convinced. \n" ); document.write( "

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