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Converse of the Parallel Lines Theorem

Date: 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 \
     ----------\-----------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!


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            /
          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:

                /  \
           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   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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