A bundle of parallels slices any two transversals in the same ratio — the engine of similarity.
Here is the bridge from numbers to shape. Take the parallel lines you already know so well and let a whole bundle of them cross two slanting roads. Like a bread slicer pushed through a loaf at an angle, they cut both roads into pieces that share the same ratio — even though the two roads have different lengths and different tilts. Pull this fact inside a triangle — draw a single line straight across it, parallel to the base — and the two sides it crosses get split in equal proportion. That one theorem, the side-splitter, is the direct source of every test for similar triangles in the next lesson. Everything about “same shape, any size” grows from this root.
In 17.1 we learned to compare two segments by their ratio and to line four segments up into a proportion a : b = c : d. Now we let parallel lines create those proportions for us, automatically, out of nothing but the geometry. We will state each fact, name it, and reason out loud — exactly the habit Stages 14–16 built.
Start with the cleanest case. Suppose three or more parallel lines are equally spaced — the same gap between each neighbour, like the rungs of a ladder. Now lay any straight line, a transversal, across all of them. Common sense says the transversal gets chopped into equal pieces, and common sense is right.
Equally spaced parallels cut equal segments. If several parallel lines are equally spaced, then every transversal that crosses them is cut into equal pieces — and this is true no matter how steeply the transversal slants.
Why is it true? Slide the “step” between the first and second parallel straight up onto the step between the second and third. The parallels are the same distance apart and run in the same direction, so the slide (a translation, from Stage 14) carries one step exactly onto the next. Matching steps means matching pieces on the transversal: the segments are congruent. A ladder makes the picture obvious — equal rungs chop both side-rails into equal lengths.
Equal spacing buys you equal pieces. Drop that condition and the pieces are no longer equal — but, as the next section shows, the ratio of the pieces is still preserved. That weaker, far more useful fact is the real prize.
Now remove the “equally spaced” restriction. Let the parallels sit at any spacing you like and cross two transversals. The pieces on one transversal are no longer equal to the pieces on the other — but they are proportional: the ratios line up perfectly.
Any bundle of parallel lines cuts two transversals into proportional segments. With the cut points named A, B, C on one transversal and A′, B′, C′ on the other,
ABBC = A′B′B′C′ and equivalently ABA′B′ = BCB′C′ .
Why? Picture squeezing many extra parallels in, equally spaced, until each original gap is sliced into a stack of thin equal strips. By 17.2.1 those strips cut both transversals into equal little pieces. If the segment AB happens to span 4 strips and BC spans 6, then on the other transversal A′B′ also spans 4 of its (equal) pieces and B′C′ spans 6 — so
AB : BC = 4 : 6 = A′B′ : B′C′. The strip-count is the same; only the strip-size differs between the two transversals.
A fan of parallels cuts the left transversal into pieces of length 4 and 6. On the right transversal the first piece measures 6. How long is the second?
Set up the proportion: 46 = 6x. Cross-multiply: 4x = 36, so x = 9. The right transversal is cut into 6 and 9 — same ratio 2 : 3, longer pieces.
Here is the move that makes everything work. Take a triangle ABC and draw a line across it parallel to the base BC, meeting side AB at D and side AC at E. That line DE splits the two sides in the same ratio.
If DE ∥ BC with D on AB and E on AC, then
ADDB = AEEC .
Equivalently AD : AB = AE : AC = DE : BC — a first taste of similarity: the little top triangle ADE is a scaled copy of the whole ABC.
Why it holds. The line DE is parallel to BC, so together they are just two members of a bundle of parallels — and the two sides AB, AC are two transversals through the common point A. By the intercept theorem of 17.2.2 the bundle cuts the transversals proportionally:
DE ∥ BC ⇒ ADDB = AEEC (intercept theorem).
In △ABC, a line DE ∥ BC cuts AB = 12 at D with AD = 8, so DB = 4 — a ratio of 8 : 4 = 2 : 1. Side AC = 9. Where does E land, and how long is DE compared to BC?
By the side-splitter, AE : EC = 2 : 1, so E divides the 9 into AE = 6 and EC = 3. And since AD : AB = 8 : 12 = ⅔, the cross-bar is two-thirds of the base: DE = ⅔ · BC.
Every good theorem has a back door. The side-splitter says parallel ⇒ proportional. Run it backward: if the sides are split in equal ratio, the cross-line must be parallel. This converse is how we prove a line is parallel using nothing but measured lengths.
If D is on AB and E is on AC with
ADDB = AEEC ,
then DE ∥ BC. Equal ratios force the line parallel — and this is precisely the lever that, in 17.3, will prove triangles similar straight from their side lengths.
Think of it as a test. Measure the four little lengths. If AD : DB matches AE : EC, stamp the line parallel ✓. If they disagree even slightly, the line tilts — it is not parallel, and extended it would meet BC’s line at some point.
The converse needs both ratios measured from the same vertex: it is AD : DB compared with AE : EC — apex-piece over base-piece on each side. Pairing AD : DB against EC : AE (upside down on the second side) is the classic slip, and it will give you a wrong verdict.
• Equal spacing → equal pieces. Equally spaced parallels cut every transversal into congruent segments.
• Any spacing → equal ratios. Any bundle of parallels cuts two transversals proportionally: AB : BC = A′B′ : B′C′ (the intercept / Thales theorem).
• Side-splitter. Inside a triangle, DE ∥ BC ⇒ AD : DB = AE : EC (and AD : AB = AE : AC = DE : BC).
• Converse. AD : DB = AE : EC ⇒ DE ∥ BC — proving lines parallel from lengths alone. Always measure both ratios from the same vertex.
Next, in 17.3, the side-splitter hands us the quick tests — AA, SAS, SSS — for when two triangles are the same shape.
Three parallel lines cut one transversal into pieces of 3 and 5. On a second transversal the matching first piece is 6. How long is the second piece?
Proportional pieces: 35 = 6x. Cross-multiply: 3x = 30, so x = 10. The pieces are 6 and 10, ratio 3 : 5 again.
In △ABC, DE ∥ BC with AD = 4, DB = 2, and AE = 6. Find EC.
Side-splitter: ADDB = AEEC, so 42 = 6EC. Then 4·EC = 12, giving EC = 3.
A line DE ∥ BC has AD : AB = 2 : 5, and the base BC = 10. Find DE.
From the side-splitter, DE : BC = AD : AB = 2 : 5. So DE = ⅖ · 10 = 4. (The little top triangle is a 2 : 5 copy of the whole.)
In △ABC, point D on AB and E on AC give AD : DB = 3 : 2 and AE : EC = 3 : 2. Is DE ∥ BC? How do you know?
Yes. The two ratios are equal — both 3 : 2, measured from the apex A on each side — so by the converse of the side-splitter, DE ∥ BC.
A ladder has equally spaced rungs. What does that tell you about how the rungs cut the two side-rails?
The rails are two transversals crossed by equally spaced parallels (the rungs), so each rail is chopped into equal pieces. Conversely, equal cuts on both rails confirm the rungs are parallel and equally spaced — the picture behind 17.2.1.
A line DE ∥ BC cuts AB so that AD : DB = 5 : 3. On the other side AE = 10. Find EC and the whole side AC.
Side-splitter: AE : EC = 5 : 3, so 10EC = 53, giving EC = 6. Then AC = AE + EC = 10 + 6 = 16.
Six questions to lock it in. Tap the answer you think is right.
This lesson is the hinge of the whole stage. Everything about “same shape, any size” — the AA, SAS, and SSS similarity tests in 17.3, the scaling laws in 17.4, the dilation in 17.6 — is built on the one theorem proved here: a line parallel to one side of a triangle divides the other two sides proportionally (and its converse). It is worth slowing down so students truly believe it before moving on.
The most common misconception is mismatched pairing. Students cheerfully write AD : DB = EC : AE — flipping the second ratio upside down — and get a wrong answer that “looks” like a proportion. Insist on a consistent rule: apex-piece over base-piece on each side, both measured from the same vertex. A second, subtler error is believing the parallels must be equally spaced to give a proportion; in fact equal spacing gives equal pieces (17.2.1), while any spacing already gives equal ratios (17.2.2). The converse widget is built so a wrong pairing produces a line that visibly tilts — let students discover that the figure refuses to lie.
Common Core. This lesson develops G-SRT.B.4 (prove that a line parallel to one side of a triangle divides the other two proportionally, and its converse), and uses G-CO.C (parallel-line and transversal reasoning) and 8.G.A (angle and parallel facts that underpin the proof). The intercept theorem is the geometric face of the proportional reasoning from 7.RP.