Introduction

C’est Mathematique! is a French expression meaning that something is so certain, it is definite. Indeed, the 19th-Century French mathematician Pierre-Simon Laplace imagined a demon that could predict every future event because it knew the exact position of every particle in the universe, and the mathematical laws that governed the motion of those particles.

Though no such demon exists, we can see why Laplace might have imagined one. Our mathematical knowledge does seem more certain than some other facts we believe to be true about politics or human nature. For instance, maybe you think that passing laws to ban handguns will reduce homicides. Then you read a study that says that actually these bans don’t have that effect. You might change your mind about whether banning handguns is a good idea… until you read another study that says that the bans do reduce accidental death among children. You’ll always have to remain open to changing your opinion based on new information.

Likewise, in the history of science, people once thought that the atom was the smallest particle that existed (the word “atom” comes from a Greek word meaning “indivisible”). Now we know that atoms are made up of protons and neutrons, and in recent decades scientists have discovered quarks. In the scientific community, people are constantly revising and improving accepted theories.

In mathematics, on the other hand, we still accept the results of ancient texts. Archimedes, Pythagoras, and Euclid are credited with many of the results that we still use today. You have probably already proven things mathematically in such a way that you’re sure they must be true. In this lesson we’ll look at what it means for a proof to be absolutely airtight — with no possibility you’ll want to revise your opinion later.

Development

Let’s think about the following two statements:

The diagonals of a parallelogram bisect each other.

The diagonals of a parallelogram bisect the angles from which they’re drawn.

When we do proofs in this section, we’re going to focus on making them airtight. We’ll use the following standard. Any time we make a statement as part of a proof, we need to be able to give a reason for that statement. That reason, in turn, needs to be a statement about geometry that we $already$ know to be true.

Here are some statements about geometry you already know to be true:

• In an isosceles triangle, the angles opposite the congruent sides are congruent.
  
  
• When parallel lines are cut by a transversal, same-side interior angles add up to 180 degrees.
  
  
• Vertical angles are equal.
  
  
• If two sides and the angle between them in one triangle are congruent to two sides and the angle between them in another triangle, then the triangles are congruent. (SAS congruence theorem)
  
  
• If the three sides of one triangle are all related to the sides of another triangle by the same proportion, then the triangles are similar. (SSS similarity theorem)
  
  
• In a parallelogram, opposite sides are congruent.
  
  

As you may recall from the first lesson of this chapter, the Greek mathematician Euclid did something very similar to what you are about to do. He boiled down his geometric knowledge to a few assumptions and then proved the rest of what he knew about geometry using those few things. He called the facts he assumed without proof “postulates.” Any fact that he had proved was a “theorem.”

Hang on to this list. You’ll be referring to it for the remainder of this chapter.

Now let’s return to the problem of proving the statements from the beginning of this lesson. If you did your job carefully, you’ll have found that the first statement was true — the diagonals of a parallelogram bisect each other.

Before writing our airtight proof of this fact, it’s important to know where to start. Since we’re proving something about a parallelogram, it’s all right to assume that we $have$ a parallelogram. So let’s draw one…

… and label the vertices so that we can talk about them later. So, specifically, $\overline {AD} \parallel \overline {BC} $ and $\overline {AB} \parallel \overline {DC} $. Those are two things we know for sure. It’s also important to be sure of what we want to show about this diagram. Let’s draw in some diagonals and call their intersection point $E$.

Then what we want to show is that $\overline {AE} \cong \overline {EC} $ and $\overline {BE} \cong \overline {ED} $.

The comic book hero G.I. Joe used to say “knowing is half the battle.” In the case of proof, knowing what information you have to start with and where you want to end up with is half the battle. Really!

What we need now is a chain of reasoning, beginning with the facts that $\overline {AD} {\kern 1pt} {\kern 1pt} \left\| {{\kern 1pt} {\kern 1pt} \overline {BC} } \right.$ and $\overline {AB} \parallel \overline {DC} $ and ending with the statement that $\overline {AE} \cong \overline {EC} $ and $\overline {BE} \cong \overline {ED} $. In this chain of reasoning, every step needs to be justified with some reason we know for sure.

When thinking about what will help us to write this proof, let’s think backwards a little bit. We want to show that certain lengths are the same. Look back at your list of postulates. What tools do you have for showing that lengths are the same? Not the results about parallel lines. Not vertical angles or linear pairs. Possibly some statements you have about parallelograms or rhombuses, but in this case the lengths we’re interested in are diagonals rather than sides of those shapes. That really only leaves one thing that could help — congruent triangles.

If we can show that certain triangles are congruent, then we can conclude that the lengths making them up are also congruent. That should be our strategy in writing this proof.

Depending on the way you answered question #5, you may have an airtight proof already. Let’s write one out as an example of a chain of reasoning in which all the steps are justified.

• Given: $\overline {AD} \parallel \overline {BC} $ and $\overline {AB} \parallel \overline {DC} $.
• Goal: Prove that $\overline {AE} \cong \overline {EC} $ and $\overline {BE} \cong \overline {ED} $.
• $\angle DAE \cong \angle BCE$ because they are alternate interior angles of the parallel lines $\overline {AD} $ and $\overline {BC} $.
  
  
• $\angle ADE \cong \angle CBE$ for the same reason.
  
  
• $\overline {AD} \cong \overline {BC} $ because opposite sides of a parallelogram are congruent.
  
  
• Therefore, $\triangle AED \cong \triangle CEB$ by the ASA congruence theorem.
  
  
• $\overline {AE} \cong \overline {EC} $ because they are corresponding parts of congruent triangles.
  
  
• $\overline {BE} \cong \overline {ED} $ for the same reason.
  
  

Are we done? YES! We’ve accomplished our goal, showing that the diagonals cut each other in half.

That proof might have been written with a little more care than you are used to. Your class will have to decide exactly how much writing is expected in a proof. But notice that every step in the argument comes from your list of postulates, so that anyone who accepts the postulates can't disagree with your conclusion — there is no “wiggle room” in this proof.

Meanwhile, you should have found that it is not true that the diagonals of a parallelogram bisect the angles. You may remember from last year that an example designed to show that something isn’t true is called a counterexample. Here’s what a counterexample to this statement might look like:

You can see from this picture that it doesn’t look true that the angles are bisected, and so it isn’t worth your time to look for a proof.

In order to start the proof that the diagonals of a parallelogram bisect each other, we had to know exactly what the word “parallelogram” meant. That’s how we knew to start with the statements “$\overline {AB} {\kern 1pt} {\kern 1pt} \left\| {{\kern 1pt} {\kern 1pt} \overline {CD} } \right.$ and $\overline {AD} \parallel \overline {BC} $”. Whenever we write a proof, we’ll have to be very precise about what we mean by the objects we’re talking about. For instance, last year you worked with the definition of a square: a quadrilateral with four congruent sides and four right angles.

Much as you can use congruent triangles to prove things about diagrams, you can also use similar triangles.

You’ve proven a couple of new things so far in this lesson. The interesting thing is, not only do you know those things, but you have proofs of them. Write the new things you know under the column labeled “theorems” on your list. Whenever you prove something new that you think may come in handy later, or that you just think is interesting or important, add it to this theorem list.

By the way, is there anything under “postulates” that you could prove and make a theorem?

Practice

In the following five problems, you should mark up the diagram with the starting information before you attempt a proof.

Given: $\overline {AB} \cong \overline {EF} $, $\angle A \cong \angle E$, $\angle C \cong \angle D$

Prove: $\overline {BC} \cong \overline {DF} $

Given: $\overline {AE} \cong \overline {CE} $, $\overline {BE} \cong \overline {ED} $

Prove: $\overline {AD} {\kern 1pt} {\kern 1pt} \left\| {{\kern 1pt} {\kern 1pt} \overline {BC} } \right.$

(In this problem and in the remaining problems in this lesson, assume that lines that appear to be straight in a figure really are straight.)

There is a joke about a mathematician asked to describe an algorithm for boiling an egg. She responds, “you take a pot, fill it with water, put in an egg, put the pot on the burner, turn the burner on, and wait 20 minutes.” “All right,” you say, but what if the pot is already sitting on the burner, filled with water?” “Then,” she says, “you take the pot off the burner and empty the water, and then you’ve reduced the problem to the previous case.”

Suppose, in the diagram below, that $\overline {PA} \cong \overline {RK} $, $\angle A \cong \angle K$
Prove: The length of $\overline {KA} $ is twice the length of $\overline {KS} $.

Now suppose instead that $\overline {PA} \cong \overline {RK} $, $\overline {PA} {\kern 1pt} {\kern 1pt} \left\| {{\kern 1pt} {\kern 1pt} \overline {KR} } \right.$ Prove: The length of $\overline {KA} $ is twice the length of $\overline {KS} $.

Given: the angles and sides as marked; W, A, and T are points on a line segment.

Prove: $m \angle T = 50^\circ$

Given: $TI = 3NI$, $RI = 3AI$

Prove: $\overline {TR} {\kern 1pt} {\kern 1pt} \left\| {{\kern 1pt} {\kern 1pt} \overline {NA} } \right.$ and $TR = 3NA$

For each figure, determine as many missing sides and angles as you can. (You don’t need to write a formal proof of your answers.) You will not be able to find $all$ of the sides and angles. Avoid using trigonometry.

In the diagram below, lines that appear parallel are parallel. Find the length of the line segment .

Going Further: Longer Proofs; Pitfalls When Writing Proofs

Sometimes the only way you can see that a statement must be true is to talk yourself through a series of steps in proof-like form. It would be hard to come to know the statement to be proven below, in any other way. In addition, the statement’s proof deals with some complications that you will sometimes come across.

Given: $\overline {XQ} {\kern 1pt} \cong \overline {XU} $ and$\angle XQY \cong \angle XUZ$.
Prove:$\angle UZY \cong \angle QYZ$.

“Working backwards and forwards” will be very important in the planning stages of this proof. At first glance, it seems that we should try to prove triangles $QYZ$ and $UZY$ congruent, since those are the triangles that contain the two angles we want to prove congruent.

In the previous problem you should have found a (shared) side and an angle, which by themselves are not enough to prove the triangles congruent. So we need to find another strategy. It may be that, using some other pair of congruent triangles in the figure, we can find that third piece of information needed to prove $\triangle QYZ \cong \triangle UZY$. Or, perhaps the solution is something completely different. Instead of doing all of our planning backwards, let’s return to thinking forwards and just look for any pair of triangles that might be congruent.

The triangles that we can actually prove congruent may have been hard to spot. As a hint, they share angle $X$. If you haven’t already, prove that these two triangles are congruent.

Now that we have congruent triangles, we can mark all of their sides and angles congruent, as follows:

One way to complete the proof is to notice that, since we now know $\triangle XYZ$ is isosceles, $\angle XZY \cong \angle XYZ$. Since these angles are both composed of two parts, and the parts $\angle XZU$ and$\angle XYQ$ are congruent to each other, that means the remaining parts must be equal as well. In other words, $\angle UZY \cong \angle QYZ$! Though this kind of part/whole reasoning doesn’t appeal to theorems or postulates that are most likely on your list, it is a common type of geometrical reasoning. Your class can decide what kind of language it wants to use for these types of arguments.

After all that, we have used congruent triangles to figure out why $\angle UZY \cong \angle QYZ$. However, much of the argument was probably done orally, in combination with a marked-up diagram. It is now time to

In the proof you just wrote, you took pains to make sure that each statement was correctly justified. However, it is easy to be sloppy when writing proofs. Two examples of proofs follow that illustrate some mistakes that are easy to make.

Consider the proof below showing that, if a quadrilateral has two pairs of congruent, opposite angles, then the quadrilateral is a parallelogram.

• First, draw a quadrilateral that can stand for any quadrilateral with congruent, opposite angles. Label the vertices $ABCD$, where $\angle A \cong \angle C$ and $\angle B \cong \angle D$.
  
  
• Given: Quadrilateral ABCD, $\angle A \cong \angle C$, $\angle B \cong \angle D$.
  
  
• Prove: $\overline {AB} \parallel \overline {CD} $, $\overline {BC} \parallel \overline {AD} $.
  
  
• Draw the diagonal $\overline {AC} $.
  
  
• $\angle B \cong \angle D$, as given.
  
  
• $\overline {AB} \cong \overline {CD} $ and $\overline {BC} \cong \overline {AD} $, because opposite sides of a parallelogram are congruent.
  
  
• Therefore, $\triangle ABC \cong \triangle CDA$ by the SAS congruence theorem.
  
  
• Then $\angle DAC \cong \angle BCA$ since they are corresponding parts of congruent triangles.
  
  
• That makes $\overline {BC} \parallel \overline {AD} $ since alternate interior angles are congruent.
  
  
• Similarly, $\angle BAC \cong \angle DCA$, again since they are corresponding parts of congruent triangles.
  
  
• So $\overline {AB} \parallel \overline {CD} $.
  
  
• Therefore, $ABCD$ is a parallelogram.
  
  

Finally, here is an example of a proof that just can’t be right: if a triangle is isosceles, then it must be equilateral. You have seen lots of isosceles triangles that are not equilateral, so you know that this proof must contain a mistake! The challenge is to find out what that mistake is. Remember that, if you agree with every step of a proof, you have no choice but to accept its conclusion.

• Given: $\triangle ABC$ with $\overline {AB} \cong \overline {AC} $.
  
  
• Prove: $\overline {AC} \cong \overline {BC} $ (which would also make $\overline {AC}$ congruent to $\overline {AB} $).
  
  
• Draw the median from $C$ and label the point where it hits $\overline {AB} $ as $D$.
  
  
• $\overline {AD} \cong \overline {DB} $ because $\overline {CD} $ is a median.
  
  
• Also, $\angle CDA \cong \angle CDB$ because, as a median, $\overline {CD} $ splits $\overline {AB} $ into two right angles.
  
  
• $\overline {CD} \cong \overline {CD} $ because any line segment is congruent to itself.
  
  
• Therefore, by SAS congruence, $\triangle CAD \cong \triangle CBD$.
  
  
• Since they are corresponding parts of congruent triangles, $\overline {AC} \cong \overline {BC} $.
  
  
• Therefore, all three sides of $\triangle ABC$ are congruent, making it equilateral.

Practice