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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 \
\
\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

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

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