Imagine you've been caught up in a twister that deposits you and your little dog in the middle of a strange new land. In order to get home, you must prove that two triangles are similar. How will you do it?
Just click your heels together three times and say, "There's no place like my geometry classroom. There's no place like my geometry classroom. There's no place like my geometry classroom."
If only!
Back here in the real world, there are three methods we can use to prove that two triangles are similar. Fortunately, they're pretty similar to a few we've already learned about. See what we did there? Aren't we a hoot?
The first time we saw these methods, they looked like helpful farmhands, tending the congruent pigs and plowing the congruent fields. Now, they're a scarecrow, a tin man, and a lion, and they're all about helping you get to the similarity wizard at the end of the Yellow Brick Road. They may look like the same theorems (in some cases, they even share the same names), but they're not the same. Those other ones were about congruent triangles, and these ones are about similar triangles.
The first method of proving similarity is the Side-Side-Side (SSS) Postulate. Here's what it says about similar triangles:
If the three sides of the two triangles are proportional in length, then the triangles are similar.
Sample Problem
Prove that these triangles are similar.
| Statement | Reason |
| 1. DO = 8, WI = 6 | Given |
| 2. OR = 12, IZ = 9 | Given |
| 3. RD = 16, ZW = 12 | Given |
4.  | Algebra |
| 5. ∆DOR ~ ∆WIZ | Side-Side-Side Postulate (4) |
The second way to prove triangle similarity is the Angle-Angle (AA) Postulate. It goes something like this:
If two triangles have two pairs of congruent angles, then the triangles are similar.
Sample Problem
Prove that these triangles are similar.
| Statement | Reason |
| 1. ∠T ≅ ∠M | Given (in figure) |
| 2. ∠TNI ≅ ∠MNA | Vertical Angles Theorem |
| 3. ∆TIN ~ ∆MAN | Angle-Angle Postulate (1, 2) |
There's one more way to prove that two triangles are similar: the Side-Angle-Side (SAS) Postulate. SAS is a nice little mash-up of AA and SSS. Kind of the way that flying monkeys are mash-ups of birds and monkeys, except the SAS is a lot more civilized and doesn't take its orders from a water-soluble witch. We think.
Basically, what it says is this:
If two side lengths are proportional and the angle in between those two sides is congruent, then the triangles are similar.
Sample Problem
Prove that these triangles are similar.
| Statement | Reason |
| 1. WI = 12, WC = 15, KE = 8, KD = 10 | Given |
2.  | Algebra |
3.  | Substitution Property |
| 4. m∠W = 35, m∠K = 35 | Given |
| 5. ∠W ≅ ∠K | Transitive Property |
| 6. ∆WIC ~ ∆KED | Side-Angle-Side (3, 5) |
Example 1
If all three sides of ∆MON are 3 units long, is it true that ∆MON ~ ∆KEY?
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Example 2
Prove that these triangles are similar.
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Example 3
Prove that ∆WTC ~ ∆WIH using the Side-Angle-Side Postulate.
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Example 4
Prove that these triangles are similar.
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Exercise 1
What is the reason for statement 1?
| Statement | Reason |
| 1. ∠B ≅ ∠K | ? |
Exercise 2
What is the reason for statement 2?
| Statement | Reason |
| 1. ∠B ≅ ∠K | Given (in figure) |
| 2. ∠BIR ≅ ∠KIC | ? |
Exercise 3
What is the reason for statement 3?
| Statement | Reason |
| 1. ∠B ≅ ∠K | Given (in figure) |
| 2. ∠BIR ≅ ∠KIC | Vertical angles theorem |
| 3. ∆BRI ~ ∆KCI | ? |
Exercise 4
What is statement 2?
Given: A, S, and E are the midpoints on their respective sides of ∆BKT.Prove: ∆BKT ~ ∆ESA
| Statement | Reason |
| 1. A is the midpoint of BK, S is the midpoint of BT, and E is the midpoint of KT | Given |
2.  | Triangle Midsegment Theorem (1) |
Exercise 5
What is the reason for statement 3?
Given: A, S, and E are the midpoints on their respective sides of ∆BKT.Prove: ∆BKT ~ ∆ESA
| Statement | Reason |
| 1. A is the midpoint of BK, S is the midpoint of BT, and E is the midpoint of KT | Given |
2.  | Triangle Midsegment Theorem (1) |
3.  | ? |
Exercise 6
What is statement 4?
Given: A, S, and E are the midpoints on their respective sides of ∆BKT.Prove: ∆BKT ~ ∆ESA
| Statement | Reason |
| 1. A is the midpoint of BK, S is the midpoint of BT, and E is the midpoint of KT | Given |
2.  | Triangle Midsegment Theorem (1) |
3.  | Triangle Midsegment Theorem (1) |
| 4. SE = ? | Triangle Midsegment Theorem (1) |
Exercise 7
What is the reason for statement 6?
Given: A, S, and E are the midpoints on their respective sides of ∆BKT.Prove: ∆BKT ~ ∆ESA
| Statement | Reason |
| 1. A is the midpoint of BK, S is the midpoint of BT, and E is the midpoint of KT | Given |
2.  | Triangle Midsegment Theorem (1) |
3.  | Triangle Midsegment Theorem (1) |
4.  | Triangle Midsegment Theorem (1) |
5.  | Statements 2, 3, 4 |
| 6. ∆BKT ~ ∆ESA | ? |
Exercise 8
Using the image below, we want to prove that ∆QET ~ ∆UES. What is the reason for statement 2?
| Statement | Reason |
| 1. QT || US | Given |
2.  | ? |
Exercise 9
Using the image below, we want to prove that ∆QET ~ ∆UES. What is statement 3?
| Statement | Reason |
| 1. QT || US | Given |
2.  | Triangle Proportionality Theorem |
| 3. ? | Reflexive Property |
Exercise 10
Using the image below, we want to prove that ∆QET ~ ∆UES. What is the reason for statement 4?
| Statement | Reason |
| 1. QT || US | Given |
| 2. | Triangle Proportionality Theorem |
| 3. ∠QET ≅ ∠UES | Reflexive Property |
| 4. ∆ QET ~ ∆UES | ? |