# Proving Segment Congruence | It All Adds Up

For necessary background information on this lesson, please click here.

Last time, we reviewed how to write a proof, and starting today, I’ll be sharing with you different purposes of writing proofs through basic geometry.

In Geometry class itself, we are about 3 chapters ahead of this, but given the importance of every postulate and theorem when writing proofs doesn’t allow for me to skip any content.

First off, let’s review what a postulate and a theorem are. You’ll see these terms many times over the coming months:

Postulate: a statement that is accepted without proof

Theorem: a statement that can be proved

Today, we’ll focus on something simple: proving line segments congruent. To start, let me teach you some new postulates and theorems:

Segment Addition Postulate: the idea that the sum of the measures of any two connected segments will equate to that of a larger angle

The Segment Addition Postulate states that segment AB + segment BC will equal segment AC.

Transitive Property: the idea that when two values are equal to each other, and another set of values are equal to each other, and one value is found in both, then the remaining two values are equal to each other

If a = b and b = c, then a = c.

Substitution Property: the idea that when one value is given to be equal to that of another value, then the two can be used interchangeably.

If a = b and a + d = c, then b + d = c.

We’ll use these, along with the other properties found in the previous edition, to write a proof. Here’s what we’re given, and what we’re told to prove:

We use those vertical tick marks to show that those two segments are congruent.

Let’s think this through. Make a plan for your proof – in this case, we are going to use the Segment Addition postulate to show that AB and BC make AC, and that it is congruent to BD. We already know that the outer segments are congruent, so we need to show that each outer segment plus the inner segment is equal to the other. In order to do this, we’ll need to show that the interior segment is congruent to itself, using the Reflexive POE. Then, using a bit of Algebra and the Substitution Property, we’ll show that AC is equal to BD.

Here is the completed proof:

We start with the given, as always. Then, like I mentioned above, we use the Reflexive Property. While we already know that the segment is equal to itself, it’s important to state that when providing a logical way of thinking. Then, we use what we know to show that each segment that we are trying to prove is made up of two smaller segments. Taking the Given, we can add the value of BC to both sides using the Addition Property of Equality. Since we know that AC = AB + BC, we can substitute AC in for it, and the same for BD.

This is just one type of proof that can be created using these properties, and I also need you to remember that your proof can vary, as long as it still makes sense.

You have everything you need to craft your own proofs using these postulates using your own sense of logic. All I can say is, think it through.

Instead of the normal Quizlet, I’ll leave you with a few problems. See if you can craft your own proofs off of them. Draw it out- that often helps me.

• In a diagram, AB = 5 and BC = 6. Prove that AC = 11. *Hint: remember that the segments are connected by B.
• Given that RT = 5 and RS = 5, and that RT = RS. Prove that RS = TS. *Hint: remember that segments can extend in opposite directions, and the first letter in a segment name indicates the starting point.
• Given that 2AB = AC, prove that AB = BC.
• CHALLENGE: M is the midpoint of AB, and N is the midpoint of CD. Prove that AM = CM. (You won’t be able to solve this based on what we learned today. You might want to do some additional research, but we’ll be covering this kind of proof next time).

In January, we’ll discuss definitions and other aspects of proofs that will be vital no matter what you are trying to prove, along with more problems to prove and helpful tips for when writing them. Until then, have a great holiday!