Thread: Parallelogram Proof...

Given: Parallelogram ABCD
BC > AB; BK = AB
Ray DK bisects angle ADC
Prove: 1) CK = CD
2) K is the midpt. of line segment of BC.

The diagram can be found on this page:
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I am really sorry for the inconvenience, but I was having problems with the file attachment. But you can fast forward the drawing speed.

let angle ADK = CDK = x. Use the fact that co-interior angles in parallel lines add to 180 to get DCK and then the angle sum of a triangle to get DKC and you are done.

Originally Posted by badgerigar

let angle ADK = CDK = x. Use the fact that co-interior angles in parallel lines add to 180 to get DCK and then the angle sum of a triangle to get DKC and you are done.

I'm sorry, but I don't understand what you mean. By the "co-interior angles" do you mean angles A and D? And DCK and DKC is the same triangle. I don't know how this can prove that CK=CD and that K is the midpt. of seg. BC.

I'm also suppose to show this as a two column proof. I was thinking that since I have to prove that CK and CD are equal, then I should first find a way to show that angles CDK and CKD are equal. That way I can prove it through the reason that "If two angles of triangle are equal in measure than the sides opp. them are equal in meas." But I don't know how to prove this. Please help. Thank you.

Nevermind I've figured it out. Thanks anyway.