Congruence, Equality, and Geometry at a Glance

One of the nice things about congruence is that it has a lot in common with equality. Since segments and angles are congruent when they have equal measures, it makes sense that congruence also has the reflexive, symmetric, and transitive properties.

In general, objects satisfying these three properties are called equivalence relations, since they behave a lot like actual equality. Martin Luther King Jr. would be proud.

So how can we take these congruent line segments and angles and convert them into proofs? Well, we'll show you.

Sample Problem

Given the following figure, show that ∠1 is congruent to ∠3.

Looking at the figure, we see that ∠1 and ∠2 are vertical angles, and ∠2 is congruent to ∠3 (by the tick marks). Thus, we can use those two facts as given statements.

Like vertical angles, you can read "betweenness" directly from a diagram. For example, in the following picture,

we can assume B is between A and C, since the segment from A to C passes through B. Like in order to drive from L.A. to Seattle, we need to pass through Oregon. We don't really have a choice, but we might as well see Portland and Crater Lake. Make the best of it.

Remember that in this situation, AB + BC = AC. Seems obvious, right? This fact is often called the Segment Addition Postulate and frequently comes up in geometrical proofs.

Here, B may not be assumed between A and C because you have to go out of your way to pass through it (like passing through Des Moines on your way from L.A. to Seattle).

Sample Problem

Given the diagram below, show that AC ≅ BD.

From the diagram, we see that AB ≅ CD by the tick marks, so that's our given information.

This problem is quite a bit trickier than what we've covered so far, since we aren't explicitly given any information about the segments we want to prove something about. So…what do we do?

We can't jump straight to working with AC and BD. Instead, let's split them into more manageable pieces. AC can be split into AB and BC, while BD can be split into BC and CD. Right away we see that BC shows up in both of them, so we can match those parts up. This leaves AB and CD to match up. Whaddayaknow? Those are exactly the segments we're given information about.

If we write it out as a proof, this is what we have.

It might seem like step 3 comes out of nowhere. Why are we adding this random BC to both sides of an equation? This is just a way to formally write out our intuitive argument about chopping big segments into smaller segments.

This idea of showing objects are congruent by splitting them into smaller pieces and matching them up is one of the most fascinating ideas in geometry. Aside from being the inspiration for several games and puzzles (like tangrams) and the backbone of many ancient proofs, this idea has led to some of the most breathtaking paradoxes in mathematics.

We've been focusing on segments for a while, and we think angles deserve some TLC, too. The Angle Addition Postulate states that the measures of two adjacent angles add up to the measure of the big angle they create. Sort of a no-brainer, but it's good to give it a name.

Some other definitions to note are: midpoints and bisectors split a segment into two congruent segments, and angle bisectors split an angle into two congruent angles. Supplementary angles add up to 180 degrees, and complementary angles add up to 90 degrees.

Again, don't worry too much about the nuts and bolts of all this. We promise we won't lead you astray.

Sample Problem

Show that if two angles are complementary to the same angle, then they're congruent.

We'll name the angles ∠1, ∠2, and ∠3. Original, we know. They're just easier to spell than names like "Bartholomew" and "Cornelius."

Given: ∠1 and ∠3 are complementary, ∠2 and ∠3 are complementary.
Prove: ∠1 ≅ ∠2

Before writing our formal proof, let's take a look at what we have. We know m∠1 + m∠3 = 90, and also that m∠2 + m∠3 = 90. Since we want to know information about m∠1 and m∠2, we can solve the two given equations for m∠1 and m∠2. This gives us m∠1 = 90 – m∠3 and m∠2 = 90 – m∠3. Now we're getting somewhere! The right sides of those equations are the same. We can now formalize this.