# Given: parallelogram and right angle; Prove: parallelogram

1. Feb 16, 2016

### Joodez

I'm curious if it looks like I defined the reasons correctly to prove that FBCD is indeed a parallelogram. Specifically, I'm unsure if I'm using #4 "definition of coincident" and #5,6 "reflexive property" correctly in their terminology. I don't have anyone else, or any teacher to ping this back to or I would try to not waste space here. This is my finished proof:

I apologize about cutting the top half of the worked problem off. It should be mirrored to the unworked problem above. If it helps, I can repost the image upon request.

2. Feb 16, 2016

### BvU

Wasn't $AD // BC$ because ABCE is a parallelogram ?

And isn't there a theorem stating $\angle F = \angle D \rightarrow FB // CD$ ?
And then you have enough to state FBCD is a parallelogram.

3. Feb 17, 2016

### Joodez

Sorry for the late reply, just got back from school.

Am I allowed to automatically assume that AD // BC because of the two 90° angles lying upon the line. I understand that this is true, but am I allowed to state this like:
AD//BC : ABCE is a parallelogram

or is there another way in which I must state this, such as:
∠F and ∠D are both _|_ (or maybe supplementary together?), whilst lying on the same line AE, which is stated to be parallel to BC based on the property of a parallelogram having parallel opposite sides.

Also, finding "parallelogram converses," I see that:

If both pairs of opposite angles of a quadrilateral are congruent, the quadrilateral is a parallelogram.

Does this reason satisfy #4. I want to say that because both of these opposite angles are congruent to one another, and also a special case of 90°, that they form a rectangle and thus must be a parallelogram, based on the definition of a rectangle.

4. Feb 17, 2016

### BvU

ABCE is a parallelogram is a given. From that you conclude $AD//BC$ so $FD//BC$.
To prove that FDBC is a parallelogram, you have to either prove $FB//DC$, or, quoting
that $\angle B = \angle D$ and $\angle C = \angle F$.

With the first you are done when there is
With the second you kind of have the reverse theorem(And I am positive that one exists -- even after 50+ years):
$AB//CE \Rightarrow \angle BAF = \angle CED$. That can lead you to $\angle BFD = \angle BCD$ and idem $\angle B = \angle D$

5. Feb 18, 2016

### Joodez

I've realized that I need to do a complete reboot of proofs. I understand most of the principles and techniques discovered in geometry. However, I cannot successfully illustrate proofs and especially the reasons for which statements are valid. Thank you for your assistance. I would like to ask one final question and that is:

What is a good resource for someone without a geometry book to learn the language and mechanics of "proofs" as utilized in geometry? So far I've been searching about for rules and etc, but if there's a comprehensive guide of any length, I would really appreciate being pointed in that direction.