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Given: parallelogram and right angle; Prove: parallelogram

  1. Feb 16, 2016 #1

    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. jcsd
  3. Feb 16, 2016 #2


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    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.
  4. Feb 17, 2016 #3
    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.
  5. Feb 17, 2016 #4


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    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##
  6. Feb 18, 2016 #5
    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.
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