Parallelogram/ triangle proof?

[stateofpeace]

Hi guys!

I'm having trouble with this problem:

Quadrilateral ABCD is a parallelogram, M is a point on AB, and N is a point on CD. If DM is perpendicular to AB and BN is perpendicular to CD, prove that triangle AMD is congruent to triangle CNB.

A diagram as to how this is supposed to look would be helpful, when I draw it, I can't get the lines perpendicular, and AMD and CNB aren't even triangles!

Thanks so much!

 

[stateofpeace]

Oh wait, oops, I just read the rules.

I honestly did try it, but it's the drawing that I can't get right. Once I get the visual down and understand how it's supposed to look, I'm pretty sure I can do the proof on my own.

Sorry about that folks!

 

[Ishuda]

Oh wait, oops, I just read the rules.

I honestly did try it, but it's the drawing that I can't get right. Once I get the visual down and understand how it's supposed to look, I'm pretty sure I can do the proof on my own.

Sorry about that folks!

Start at point A and go right to point B. From B go at 45 degrees or so down (about 'half down') 'a bit' to point C. From C, go left, parallel to AB, to point D as much as you went right from point A to point B. Point M is directly above point D and point N is directly below point B.

Looks like a SAS should work for congruence.

 

[stateofpeace]

Thanks, I drew it out with your guidance! It basically creates a rectangle in the center, with two triangles that can easily be proven congruent at the two ends! Now it seems so obvious!

Did you automatically know how to draw it? Or did you have to try it a few times?

 

[Ishuda]

Thanks, I drew it out with your guidance! It basically creates a rectangle in the center, with two triangles that can easily be proven congruent at the two ends! Now it seems so obvious!

Did you automatically know how to draw it? Or did you have to try it a few times?

Had to draw it twice. First just drew a parallelogram (went 'backwards' at B) and saw what was wrong.