How this instrument works
The Triangle Proportionality Theorem says a single fact with wide reach: draw a line parallel to one side of a triangle, crossing the other two sides, and it divides both of those sides in exactly the same ratio — a₁ ⁄ a₂ = b₁ ⁄ b₂. Euclid stated it as Proposition 2 of Book VI of the Elements, and the reasoning behind it is short: the parallel line closes off a smaller triangle at the vertex that shares the original angle there, and because parallel lines cut any transversal at matching corresponding angles, that smaller triangle shares a second angle too. Two equal angles force AA similarity, and similar triangles carry every side ratio in lockstep, which is the proof in outline.
Set a₁ equal to a₂ and the proportion collapses to b₁ = b₂ as well — the parallel line then crosses the exact midpoint of both sides at once, the special case handled directly by this site's Midsegment of a Triangle calculator. The theorem also runs in reverse: if a line crossing two sides produces matching ratios on both, that line is guaranteed parallel to the third side, a test worth reaching for when an angle is awkward to measure directly but a length is not.
The theorem says nothing about the segments' absolute lengths, only their ratio, so it scales without limit — a blueprint at 1:100 and the finished building both obey the identical proportion. The one place it goes wrong in practice is segment pairing: a₁ and b₁ must both lie between the vertex and the parallel line, while a₂ and b₂ must both lie between that line and the base; swap a top segment for a bottom one and the cross-multiplication still produces a number, just not one that describes the triangle in front of you.
- Enter your triangle's first side into Segment a₁ (top, side 1) and Segment a₂ (bottom, side 1) — a₁ nearer the vertex, a₂ nearer the base the parallel line does not touch.
- Enter the matching segment on the second side into Segment b₁ (top, side 2), keeping it on the same side of the parallel line as a₁.
- Leave Segment b₂ (bottom, side 2) empty — it is the value the sheet solves for.
- Read Segment b₂ (bottom, side 2): the sheet applies b₂ = a₂ × b₁ ⁄ a₁ so that a₁ ⁄ a₂ equals b₁ ⁄ b₂.
- Double-check the pairing before trusting the result — a₁ and b₁ should both sit between the vertex and the parallel line, and a₂ and b₂ both between that line and the base.
Worked example — segments 4, 6, and 6
Picture a triangle with a line drawn parallel to its base, cutting the other two sides into a near segment and a far segment each. On the first side, the vertex-side segment (Segment a₁) measures 4 units and the base-side segment (Segment a₂) measures 6 units. On the second side, the vertex-side segment (Segment b₁) measures 6 units. Because a parallel line always divides the two sides it crosses in the same ratio, a₁ ⁄ a₂ must equal b₁ ⁄ b₂ — cross-multiplying gives b₂ = a₂ × b₁ ⁄ a₁ = 6 × 6 ⁄ 4 = 9.0 units for Segment b₂, the base-side segment on the second side.
Check the arithmetic by comparing ratios directly: 4 ⁄ 6 reduces to 2 ⁄ 3, and 6 ⁄ 9 reduces to that same 2 ⁄ 3, confirming the two sides really are cut proportionally rather than by coincidence. That equality is also a working test in the other direction — if a measured line across a triangle produces segments in matching ratios like these, the theorem's converse guarantees the line is parallel to the third side, no protractor required.
Questions
What does the Triangle Proportionality Theorem actually state?
If a line runs parallel to one side of a triangle and crosses the other two sides, it cuts those two sides in the same ratio: a₁ ⁄ a₂ = b₁ ⁄ b₂, where a₁ and b₁ are the segments nearer the vertex and a₂ and b₂ are the segments nearer the base. It is also called the side-splitter theorem, since the parallel line splits both sides at matching proportions.
Why does a parallel line force the same ratio on both sides?
The parallel line creates a smaller triangle at the vertex that shares the original triangle's angle there, and because parallel lines cut a transversal at equal corresponding angles, the two triangles share a second angle too. AA similarity then locks every pair of corresponding sides into one constant ratio, which is exactly a₁ ⁄ a₂ = b₁ ⁄ b₂.
How is this related to the Triangle Midsegment Theorem?
The midsegment case is this theorem at a ratio of exactly 1:1 — when a₁ equals a₂, the proportion forces b₁ to equal b₂ too, so the parallel line crosses the midpoint of both sides at once. That special case, where the cutting segment also connects two midpoints, is covered separately by this site's Midsegment of a Triangle calculator.
Does the theorem work in reverse, to test whether a line is parallel?
Yes — the converse holds: if a line crosses two sides of a triangle and divides them in the same ratio, that line must be parallel to the third side, with no angle measurement required. Work out the two ratios first; if a₁ ⁄ a₂ and b₁ ⁄ b₂ match, parallelism is confirmed.
What is the most common mistake when applying this theorem?
Pairing the wrong segments — matching a vertex-side segment on one side with a base-side segment on the other instead of like with like. a₁ and b₁ must both sit between the vertex and the parallel line; a₂ and b₂ must both sit between the parallel line and the base. Swap that pairing and the cross-multiplication still returns a number, just not the right one.
Does the theorem still work if the parallel line falls outside the triangle?
The version here assumes the line crosses between the vertex and the base, producing two positive segments on each side. A parallel line extended outside the triangle needs a signed, extended form of the same ratio, which is a different setup from the four positive segment lengths this sheet expects.