Angles and Sides

Are you tired of hearing that similar triangles have congruent angles and proportional sides? We only keep saying it because it's important. And true. And important.

If we know that two triangles have congruent angles and proportional sides, then we know that they're similar. Actually, if we know that two triangles have congruent angles and/or proportional sides, then we know that they're similar. We don't have to know both, though it is nice when you've got all your bases covered.

Check out these triangles.

They sure do look similar, but sometimes drawings can throw us a curveball. Fortunately, the triangles' side lengths are given to us, which is handy, because we can't tell just from looking whether the angles are congruent.

Let's set up the ratios and compare. Are they equal?

Yes, indeed, the side lengths are proportional, so we can be sure that ∆OAK ~ ∆MIL.

Sample Problem

∆PIT has side lengths 4, 5, and 8, while ∆KAN has side lengths 12, 15, and 22. Are they similar?

Let's set up some ratios and see.

Nope. So they're not similar.

It's a whole 'nother ball game when we're dealing with angles. While side lengths are proportional in similar triangles, the angles have to be congruent. If we cut out any two similar triangles (which we wouldn't recommend doing to your screen, but feel free to print the page and then proceed with the paper) and placed them on top of each other, we could line up each set of corresponding angles and see that the angles are exactly the same size.

Triangles ∆TEX and ∆WAS are similar because their angles are congruent. Specifically, ∠T ≅ ∠W, ∠E ≅ ∠A, and ∠X ≅ ∠S.

Sample Problem

∆NYM has angles that measure 50°, 87°, and 43°. ∆PHI has angles that measure 87°, 50°, and 43°. Are they similar?

Yep. Mark it down: a triple play with three pairs of congruent angles. That makes the triangles similar.

This begs an interesting question. Do we have to know the measures of all the angles in order to decide whether two triangles are similar?

In short, no.

The long answer relies on the fact that the three angles inside every triangle always add up to 180° (remember the Angle Sum Theorem for Triangles?). That means that if we only know the measure of two of the angles, we can figure out the measure of the third one. Ballpark figures won't cut it, though.

So what about these two triangles? Are they similar?

In the first triangle, we know ∠P = 48° and ∠I = 79°. Since m∠P + m∠I + m∠T = 180, we can calculate that m∠T = 53°. In the second triangle, m∠L = 48° and m∠A = 79°. Again, since m∠L + m∠A + m∠D = 180, we know that ∠D must be 53° as well. Both triangles have three congruent angles, which means they must be similar.

To write the similarity statement, we have to get our letters in the right order. Congruent angles must be in the same place in each triangle's name: ∠P ≅ ∠L, ∠I ≅ ∠A, and ∠T ≅ ∠D, so we write ∆PIT ~ ∆LAD.

But we really didn't have to calculate the size of those third angles. Here's why:

Even without needing the exact measurements of any of the angles, we know that ∠T and ∠D are equal in measure, meaning the that they're also congruent.

And we've just discovered something called the Angle-Angle Postulate, which says that if two triangles share two pairs of congruent angles, then the triangles are similar. See, isn't sharing nice?

Sample Problem

Two of the angles in ∆BOS measure 22° and 108°, while two angles in ∆CLE measure 108° and 48°. Are the triangles similar?

We know from the problem that they have at least one congruent angle (108°). But do they have another set? To find the missing angle in ∆BOS, we'll solve 180° – (108° + 22°) = x, or x = 50°. The three angles in ∆BOS are 22°, 50°, and 108°. None of them is 48°, so we already know (without figuring out the size of ∆CLE's third angle) that our two triangles don't have at least two congruent angles. They don't match, so the two triangles are not similar. A swing and a miss.