Same shape, any size — and three quick tests (AA, SAS, SSS) to prove it.
Zoom in on a photo and it gets bigger, but it never changes shape: the angles hold, and every length stretches by the same factor. Two triangles like that are similar, written △ABC ~ △A′B′C′. The lovely news is that we never have to check all six things — three angles and three sides. The side-splitter theorem from the last lesson hands us three quick tests: just two angles (AA), or two sides and the angle between them (SAS), or all three sides in proportion (SSS). And a bonus gift hides inside every right triangle: drop the altitude to the hypotenuse and three similar triangles fall out, carrying the elegant geometric-mean relations.
Two triangles are similar when one is a faithfully scaled copy of the other. We write it with a tilde and read the letters in order:
△ABC ~ △A′B′C′
The statement packs in two promises at once. First, corresponding angles are equal:
∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′.
Second, corresponding sides are proportional — they all share one ratio, the scale factor k:
A′B′AB = B′C′BC = C′A′CA = k.
The letter order in △ABC ~ △A′B′C′ tells you which vertex matches which: A↔A′, B↔B′, C↔C′. So △ABC ~ △DEF is a different statement from △ABC ~ △EFD — the second pairs A with E, B with F, C with D. Always write the names so the correspondence reads straight off.
Notice what happens when the scale factor is exactly k = 1: every side of the copy equals its partner, so the two triangles are congruent. Congruence is just similarity that has stopped resizing — the special case k = 1. Everything you proved about congruent triangles in Stage 15 is the no-stretch corner of the bigger world we are entering now.
Six facts is a lot to verify. Here is the first shortcut, and the one you will reach for most often. If two pairs of angles match, the triangles are similar — full stop:
if ∠A = ∠A′ and ∠B = ∠B′, then △ABC ~ △A′B′C′ (AA).
Why is two enough? Because the three angles of any triangle add to 180°. If two pairs already agree, the third pair has no choice — ∠C = 180° − ∠A − ∠B must equal ∠C′ = 180° − ∠A′ − ∠B′. Pin down two and the third is forced. That is why the test is called AA rather than AAA.
Slide the small triangle into the big one so vertex A sits on A′ and side AB lies along A′B′. Because ∠B = ∠B′, side BC comes out parallel to B′C′. But a line parallel to one side of a triangle splits the other two sides proportionally — the side-splitter theorem from 17.2. So the sides are in proportion automatically. Equal angles drag proportional sides along behind them.
The picture below builds two triangles from the same two angles but at different sizes. Drag the angles and watch: the shapes track each other perfectly, no matter how far apart their sizes drift.
Most similarity problems are solved with AA, because angles are so easy to find: parallel lines hand you equal corresponding angles, a shared angle is equal to itself, two right angles match instantly. Hunt for two equal angles first; only if angles run out do you reach for SAS or SSS.
Sometimes you know sides, not angles. The second test mirrors the SAS congruence rule from Stage 15, with one word changed — equal becomes proportional:
if A′B′AB = A′C′AC and the included angle ∠A = ∠A′, then △ABC ~ △A′B′C′ (SAS).
The word included is the whole story: the equal angle must sit between the two proportional sides, hugged on both sides by them. SAS congruence asks for the two sides to be equal; SAS similarity relaxes that to in the same ratio.
Triangle one has sides 4 and 6 around a 50° angle. Triangle two has sides 6 and 9 around a 50° angle. Check the ratios: 64 = 1.5 and 96 = 1.5 — the same. Two sides in ratio 3 : 2, the included angle equal ⇒ the triangles are similar by SAS, with k = 1.5.
If the equal angle is not tucked between the two proportional sides, the test fails. Two proportional sides with an equal angle off to the side — call it SSA — does not prove similarity. We will see that trap break in the next widget.
The third test uses no angles at all. If all three pairs of sides are in the same ratio, the triangles are similar; the angles then take care of themselves:
if A′B′AB = B′C′BC = C′A′CA, then △ABC ~ △A′B′C′ (SSS).
Triangle one is the famous 3–4–5. Triangle two is 6–8–10. Line the sides up: 63 = 84 = 105 = 2. All three ratios equal 2, so △ ~ △ by SSS, with k = 2.
The three ratios must all be equal. Compare 3–4–5 with 6–8–11: now 63 = 2 and 84 = 2, but 115 = 2.2 ≠ 2. Two out of three is not enough — those triangles are not similar.
The widget below lets you name the test for yourself. Switch between the three valid tests — and one famous fake.
| Test | What you check | Verdict |
|---|---|---|
| AA | two pairs of angles equal | ~ similar |
| SAS | two sides proportional + included angle equal | ~ similar |
| SSS | all three sides proportional | ~ similar |
| SSA | two sides proportional + a non-included angle | ✗ not a test |
Now a small miracle that has been waiting inside every right triangle you have ever drawn. Take △ABC with its right angle at C, and drop the altitude CD from C straight down to the hypotenuse AB. That single line splits the big triangle into two smaller ones — and astonishingly, all three are the same shape:
△ACD ~ △CBD ~ △ABC.
Why? Look at the little triangle △ACD. It shares angle ∠A with the big triangle, and it has its own right angle at D. Two equal angles — that is AA. The same argument works for △CBD, which shares ∠B. Both small triangles are similar to the big one, so they are similar to each other.
Every point on a semicircle drawn over AB sees that diameter at a right angle — a fact we will prove properly in Stage 18 (circles). That is the cleanest way to keep ∠C = 90° exactly while C moves: the figure can never lie about the right angle.
Those three similar triangles are not just pretty — they hand us three of the most elegant length relations in all of geometry. Label the pieces the altitude makes on the hypotenuse: let p = AD and q = DB, so the whole hypotenuse is c = p + q, and call the altitude h = CD. Matching corresponding sides in the similar triangles gives:
h² = p · q · AC² = p · c · CB² = q · c
Read them out loud. The altitude h is the geometric mean of the two hypotenuse pieces. Each leg is the geometric mean of the whole hypotenuse and the piece next to it.
The geometric mean of two positive numbers m and n is √(m·n). So h² = p·q says exactly h = √(p·q): the altitude is the geometric mean of the two pieces it stands between. (These are the seeds of the "intersecting-chords" relations you will meet in Stage 18.)
Take legs 3 and 4, hypotenuse c = 5. The altitude to the hypotenuse is h = (3·4)/5 = 2.4. The pieces are p = 9/5 = 1.8 and q = 16/5 = 3.2. Check the relations: h² = 5.76 = 1.8 · 3.2 ✓, and AC² = 9 = 1.8 · 5 ✓. Switch the widget above to Geometric means and watch the numbers land on these identities exactly.
Similarity means same shape, any size: equal corresponding angles, proportional corresponding sides, written △ABC ~ △A′B′C′ with the letters matched in order. To prove it you need only one of three tests.
| If you know… | Test | Conclusion |
|---|---|---|
| two equal angles | AA | ~ similar |
| two proportional sides + the included angle | SAS | ~ similar |
| three proportional sides | SSS | ~ similar |
Drop the altitude from the right angle of a right triangle to its hypotenuse and you get three similar triangles in one, and with them the geometric-mean relations: h² = p·q, AC² = p·c, CB² = q·c. Congruence is similarity with k = 1. Next lesson collects the rewards — every length scales by k, while area scales by k².
△ABC ~ △DEF with ∠A = 40° and ∠B = 75°. Find ∠F.
Corresponding angles are equal, so ∠F = ∠C. And ∠C = 180° − 40° − 75° = 65°.
△ABC ~ △DEF with scale factor k = 3/2 and AB = 8. Find DE.
DE corresponds to AB, so DE = k · AB = (3/2) × 8 = 12.
Are triangles with sides 6–8–10 and 9–12–15 similar? If so, name the test.
Check all three ratios: 9/6 = 12/8 = 15/10 = 3/2. All equal, so yes — similar by SSS, with k = 3/2.
A right triangle has legs 6 and 8 and hypotenuse 10. Find the altitude to the hypotenuse, and the two pieces it cuts.
Altitude h = (leg · leg)/hyp = (6·8)/10 = 4.8. The pieces are p = 6²/10 = 3.6 and q = 8²/10 = 6.4 (and 3.6 + 6.4 = 10 ✓; check h² = 23.04 = 3.6 · 6.4 ✓).
The altitude to the hypotenuse is the geometric mean of the two segments it makes, here 4 and 9. Find the altitude.
h = √(p·q) = √(4 × 9) = √36 = 6.
Explain why the AA test needs only two equal angles, not three.
The three angles of any triangle sum to 180°. Once two pairs agree, the third pair is forced: ∠C = 180° − ∠A − ∠B must equal ∠C′ = 180° − ∠A′ − ∠B′. So checking the third angle is redundant — two pin down the shape.
Six questions to lock it in. Tap the answer you think is right.
The big idea is economy: similarity demands six matchings (three angles, three sides), but you never have to confirm all six. Just as congruence has SSS / SAS / ASA, similarity has AA, SAS, and SSS, and each follows from the side-splitter theorem of the previous lesson. AA is by far the most used — steer students to hunt for two equal angles first, since parallel lines, shared angles, and right angles supply them everywhere.
The misconception to watch is the SSA trap: students assume that two proportional sides plus any equal angle proves similarity. It does not — the equal angle must be the included one (between the two sides). The "Name the test" widget shows two genuinely different triangles that share two proportional sides and a non-included angle, so the failure is visible, not merely asserted. Two related slips: mismatching the correspondence in △ABC ~ △DEF (the letter order is data, not decoration), and confusing the altitude relation h² = p·q with h = p·q — it is the geometric mean, a square root, not the product.
Common Core: this lesson supports G-SRT.A.3 (use the properties of similarity transformations to establish the AA criterion), G-SRT.B.4 and G-SRT.B.5 (prove and use the AA / SAS / SSS similarity criteria, and the geometric-mean relations in a right triangle to solve problems).