[G.CO.10] Congruence #10

Objective

Common Core Text:

• [G.CO.10] Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

Said Differently:

• • Prove the following theorems about triangles

    • Triangle Sum Theorem

    • Isosceles Triangle Theorem

    • Midsegment Theorem

    • Concurrency of Medians Theorem

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Explanation

Angle Addition Postulate

• If you add the two angles that make a bigger angle, you get the bigger angle.

It's the same thing as saying:

• If you divide a whole into parts

• then add the parts together

• you get the whole.

Remember, a postulate is something that is supposed to be so easy that everyone just assumes it is true. You just have to believe this one. This postulate is used within other proofs

Triangle Sum Theorem

• The interior angles of a triangle sum to 180°

An extremely useful theorem. Comes up all the time. Let's do an informal proof to understand, then we'll do a 2 column proof.

Here we have triangle ΔABC, with angles ∠1, ∠2, and ∠3.

Let's draw a line parallel to line segment AB. We'll call our new line L.

This creates two new angles, ∠4 and ∠5.

Notice that now we have the classic "two parallel lines and a transversal". That means we can use our angle theorems; namely, we can see that ∠1 and ∠4 are alternate interior angles, so they must be congruent.

I've changed the color on ∠4 to show this

Similarly, we can see that ∠2 and ∠5 are alternate interior angles, and therefore congruent

I've changed the color of ∠5 to show this

Here comes the juicy part. Look at ∠4, ∠3, and∠5. You can see that if you put them all together, you get a straight line (line L). In other words, using our Angle Addition Postulate, m∠4 + m∠3 + m∠5 = 180°. But since ∠4 ≅ ∠1 and ∠5 ≅ ∠2, that's the same thing as saying m∠1 + m∠3 + m∠2 = 180°. Hey, that's what we were trying to prove! The interior angles of a triangle sum to 180°.

Here's a more formal, 2-column proof

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Prove that the sum of the interior angles of a triangle equals 180°.

Isosceles Triangle Theorem

• The base angles of an isosceles triangle are congruent.

Also very useful.

Let's do an informal proof to understand, then we'll do a 2 column proof.

Here's an isosceles triangle. We need to prove that angle ∠1 ≅ ∠2. There are many ways to do this. Of course, Euclid has his own proof, but my favorite is Legendre's.

Let's draw a point D that bisects line AB. Then, lets connect C and D with a line segment.

Well what have we here. Two triangles, AD = DB, AC = CB, and of course, CD = CD. All three sides are congruent, so by SSS, the two triangles are congruent. At this point, we can use CPCTC to say that ∠1 = ∠2 and viola, we're done.

Here's a more formal, 2-column proof

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Prove that the base angles of an isosceles triangle are congruent.

Midsegment Theorem

• The segment joining midpoints of two sides of a triangle is parallel to the third side and half the length.

For this picture, the midpoint theorem says that

1. Line DE || BC (are parallel)

2. Length of DE = ½BC

Let's see if we can prove this

Well since we know midpoints split a line exactly in half, we know that

• AB ÷ AD = 2 (AB is twice the length of AD)

• AC ÷ AE = 2 (AC is twice the length of AE)

Since these two ratios are the same, and they have an included angle ∠1, we have SAS for similar triangles (see [G.SRT.2] for more about similar triangles). Since all corresponding sides have the same ratio

• BC ÷ DE = 2 (BC is twice the length of DE)

which is the same thing as

• • DE = ½BC (DE is half the length of BC) (Hey we just proved #2)

Now comes the tricky part. I'm going to extend a few lines and erase some things to make it easier to see what I'm talking about. This will be OK because for the next part we're only worried about angles, and I'm not going to change any angles.

Ok. Look familiar? It's the classic "two parallel lines and a transversal"... or is it? We actually don't know if DE and BC are parallel yet. That's what we're trying to prove.

According to the Corresponding Angles Theorem, we know that if DE and BC were parallel, then ∠2 and ∠3 would be corresponding angles. And if they are corresponding angles, then they are congruent. Let's make a little picture of that logic.

• If DE || BC → ∠2 and ∠3 are corresponding angles → ∠2 ≅ ∠3

But wait, we said that these two triangles are similar. Corresponding angles of similar triangles are congruent. That means we already know that ∠2 and ∠3 are congruent! That means we can work backwards instead. This would be called the Converse of the Corresponding Angles Theorem. Notice the arrows in the logic change direction.

• DE || BC ← ∠2 and ∠3 are corresponding angles ← If∠2 ≅ ∠3

And now we know that DE || BC (Hey, we just proved #1)

Here's a more formal, 2-column proof

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Prove that the segment joining midpoints of two sides of a triangle is

1. parallel to the third side and

2. half the length.

Concurrency of Medians Theorem

This is a very challenging theorem to prove, and doesn't really seem appropriate for Grade 10 Geometry. If you would like to see the proof it is here.

http://jwilson.coe.uga.edu/EMAT6680/Brink/allasgnmts/Graphs4/concurr3.html

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