Converse of the Parallel Lines TheoremDate: 05/28/2000 at 17:55:51 From: Sarah Subject: Converse Proofs Dr. Math, I'm homeschooled and I am really having trouble with this "converse proof" question. "Use the following diagram to prove the converse of the Parallel Lines Theorem: If a transversal intersects two lines so that the alternate angles are equal, then the lines are parallel." \ ------\---------------L1 Given: <x = <y x \ \ \y ----------\-----------L2 Prove: L1 is parallel to L2 \ I read quite a few of your FAQs, and all of them dealt with two-column proofs. That is not the way they have us do it. Instead, we are to assume that the theorem is incorrect (in this case, the angles are not equal), and disprove the assumption, therefore proving the theorem is correct. Another question I am having trouble with is "In triangle DEF, G and H are the midpoints of DE and DF, respectively. The line segments EH and FG are medians. Prove that these medians cannot bisect each other." There is a diagram, but I can't really draw it. Unfortunately, all that the answer part of my text has for these questions is "answers may vary." Please help! Thanks, Sarah Date: 05/28/2000 at 23:36:48 From: Doctor Peterson Subject: Re: Converse Proofs Hi, Sarah. First, let me make a comment about two-column proofs. Their prevalence in our FAQ reflects the fact that most American schools seem to teach this way to present a proof - or maybe it's that most students who are confused about proofs are taught that way. Mathematicians generally present proofs in paragraph form, and I've heard of some being baffled when they first saw a two-column proof, which seemed to them overly rigid. It has value, however, in introducing a student to the concept of careful proof, sort of like training wheels for the mind. On the other hand, I don't think the form is related to the difference between direct and indirect proof; an indirect proof can be written in either form, though the two-column form may be better adapted to direct proofs. Let's work through this. You'll notice that your description of what you had to do was a little off; you have to assume the conclusion is false, and prove that the given must be false. Given: <x = <y To prove: L1 || L2 \ /L2 \ A / ------+-----------+------------L1 x \ z / C \ / \y / + / B\ / \ Assume that the conclusion is false: L1 is NOT parallel to L2. Then L1 meets L2 at some point C, forming a triangle ABC. What contradiction arises? Here's one: Looking at triangle ABC, we know then that <x = <y + <z; but since <z > 0, this implies that <x is not equal to <y, and therefore that they are not congruent. This contradiction proves the theorem. The details will depend on what theorems you have available; if they haven't proved the external angle theorem I used, then there will be something similar you can use in its place. But however you fill in the details, the idea is to draw a (wrong) diagram in which the conclusion is false, and then use it to prove the premises are false. Let's look at your second problem: In triangle DEF, G and H are the midpoints of DE and DF, respectively. The line segments EH and FG are medians. Prove that these medians cannot bisect each other. Here's my attempt at a diagram: D + / \ G / \ H + + / + \ / / X \ \ +-----------------+ E F This sounds odd to me, because I know a theorem that says the medians divide one another in the ratio 1:2, so I could just state that theorem and then say 1:2 is not that same as 1:1, so it doesn't bisect. Given that, this is a silly theorem, since there's a much stronger positive statement to be made. I'll assume you haven't learned that yet, so you can't use it. In that case, this sounds like a time to use proof by contradiction again, since the conclusion has a negative form. Let's suppose that EH and FG intersect at point X which is their midpoint. Then EX = XH. What can we look for that would lead to a contradiction with the assumption that G and H are midpoints of DE and DF? I look at the figure and recognize that GH is parallel to EF, and alternate interior angles tell me that GHX and FEX are similar triangles. But if EX = XH, then they are actually congruent triangles! See if you can use that to complete the proof. Or, you might find that you can rework this as a direct proof that EX > XH. But if you do that well, you'll find that it actually proves that EX = 2XH, the theorem I mentioned before. I can understand your difficulty with the lack of answers to questions like this; proofs are hard to make entirely self-teaching even if the texts are designed for that (which yours clearly isn't), since they're more like writing an essay than the more straightforward parts of math. My homeschooled son is fortunate to have me available for this sort of thing, and that's one of the reasons I'm a math doctor, so others can get help like this. Feel free to write again! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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