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Date: 11/09/1999 at 22:15:30
From: Amy DeMent
Subject: Two-Column Proofs

I am in geometry, and I am completely lost on proofs. I understand you
have some givens, and you have to prove something, but as to the steps
in between, I am clueless. If I have one side of the proof I can get
the other side, or if I am looking at a completed proof I can see how
it was done, but I don't understand how to come up with the
statements.

For instance, here is an example that I copied down from the board
that I don't understand. The only reason it is completed is because it
was done for me on the board.

Given: AP = BP
AQ = BQ

Prove: PQ is perpendicular to AB

P
/| \
/ |  \
A /__R___\B
\  |   /
\ |  /
\| /
Q

PROOF
Statements                         Reasons
----------                         -------
1. AP = BP                         1. Given
2. AQ = BQ                         2. Given
3. PQ = PQ                         3. Reflexive property
4. triangle PAQ = triangle PBQ     4. SSS
5. <APR = <BPR                     5. CPCTC*
6. PR = PR                         6. Reflexive property
7. triangle PRA = triangle PRB     7. SAS
8. m<PRA = m<PRB                   8. CPCTC*
9. <PRA, <PRB are linear pair      9. If alt. int. lines are
congruent, the lines are
parallel
10. m<PRA = m<PRB = 90            10. If 2 <'s are congruent and
form a linear pair, each is
a right angle
11. PQ is perpendicular to AB     11. Def. of perpendicular lines

*CPCTC = Corresponding parts of congruent triangles are congruent

And that is the whole proof. Could you please explain why each step is
done and how you came up with it?

Amy
```

```
Date: 11/10/1999 at 09:17:26
From: Doctor Peterson
Subject: Re: Two-Column Proofs

Hi, Amy. Thanks for a well-written and thorough question.

I'd recommend looking through our FAQ on proofs:

http://mathforum.org/dr.math/faq/faq.proof.html

This is a compilation of previous answers to questions, and will show
you both examples of how we've done simple proofs, and explanations of
ways to think about it. I'll deal with your particular example, but
there you will find several different ways to approach proofs that
will be useful to you.

>     Given: AP = BP
>            AQ = BQ
>
>     Prove: PQ is perpendicular to AB
>
>          P
>         /| \
>        / |  \
>     A /__R___\B
>       \  |   /
>        \ |  /
>         \| /
>          Q
>
>                        PROOF
>     Statements                         Reasons
>     ----------                         -------
>     1. AP = BP                         1. Given
>     2. AQ = BQ                         2. Given

Having written down the givens, and without having looked at the rest
of the proof yet, let's think about what we have and where we want to
get to - look over the territory before we start marching across it.

We have a kite shape, and want to prove that the "sticks" are
perpendicular. My first thought is that they're held that way because
the string around the edge of the kite pulls equally in both
directions and holds them straight. That has nothing to do with
geometry, but maybe it can give me a feel for how this will work.

Then I look at which edges are equal, and realize that we have two
isosceles triangles, APB and AQB. PQ, if the conclusion is right, will
be an altitude of both triangles.

Next, looking ahead to our goal, I look for some possible congruent
triangles. APB and AQB are not going to be congruent, but within each
of them the altitude ought to form two congruent triangles, APR = BPR
and AQR = BQR. If I can prove these, maybe I'll be going in the right
direction.

Now let's see what comes next...

>     3. PQ = PQ                         3. Reflexive property
>     4. triangle PAQ = triangle PBQ     4. SSS

Ah! They thought of something better: each side of the whole kite will
be congruent. This is essentially because the kite is symmetrical: the
sides are the same because opposite sides match. We stated PQ = PQ
just to get all three sides so we can apply the SSS theorem.

Now where will this get us? It says nothing yet about the angles at O;
yet it seems related to the goal because if we fold the kite over
along line PQ, the fact that A and B will meet says the line between
them is perpendicular to the fold (or "mirror"). How do we show that?

>     5. <APR = <BPR                     5. CPCTC*
>     6. PR = PR                         6. Reflexive property
>     7. triangle PRA = triangle PRB     7. SAS

Just what I was thinking: the congruence of PAQ and PBQ, which we
proved on the basis of their sides, tells us something about their
angles, which we can use in other triangles. Here we've shown that the
top two triangles are congruent. Remember how I said the sticks in a
kite are held perpendicular because they're pulled equally from each
side? That's what we're seeing now: angles ARP and BRP will be
congruent because the two sides are the same, and two congruent angles
that are supplementary have to be right angles. Let's see if that's
how they finish this.

>     8. m<PRA = m<PRB                   8. CPCTC*
>     9. <PRA, <PRB are linear pair      9. If alt. int. lines are
>                                             congruent, the lines are
>                                             parallel
>     10. m<PRA = m<PRB = 90            10. If 2 <'s are congruent and
>                                            form a linear pair, each
>                                            is a right angle
>     11. PQ is perpendicular to AB     11. Def. of perpendicular
lines

Yep, I was right. So we're done.

Let's look back over what we did. If I were doing this on my own,
rather than trying to follow (and sometimes anticipate) someone else's
proof, I would have thought something like this:

I'm given some equal sides; from that I may be able to prove some
triangles congruent. At the other end, I want to come out with a
statement about angles, and since right angles are congruent to one
another, that can probably come from some congruent triangles.

I'd work from both ends: the triangles I can prove congruent from the
givens are PAQ and PBQ; the triangles I'd want to prove congruent in
order to get to my conclusion are either PRA and PRB or QRA and QRB (I
could do the proof using either pair.) Then I'd look for a connection
between PAQ and PRA, and would see that they share an angle and a
side. At that point I'd be almost done.

As I explain in some of my answers in the FAQ, this illustrates that
there can be a lot of looking around before we hit on the actual path
to a proof, and that a proof can be likened to building a bridge,
starting from both shores and meeting in the middle. There may also be
some wrong turns along the way, or some good ideas that aren't needed
for this particular proof - but that might give you ideas for another.

Also, you don't have to think of the statements one at a time. They
come in groups: prove these triangles congruent, then prove those
triangles congruent. Get the main idea, then work out how to say the
details.

I hope this helps you out. Learning proofs can be a challenge,
because it's a new idea to most students, and a new way of thinking;
but if you can use your experience in other kinds of thinking, you can
do fine.

Let me know if you need any more help.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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