3e · Geometry

Recognising when to use Thales

The configurations to spot before calculating, and the most common proportion mistake.

AiCole teaching team3 min read · October 14, 2026

Two lines meeting at A, cut by two parallel lines: the Thales configuration.

Thales’ theorem is one of those tools students know by heart… and use at the wrong moment. On a test paper, the difficulty is almost never the calculation: it is recognising the figure, checking the right condition and writing the ratios in the right order.

The idea in one sentence

When two parallel lines cut two intersecting lines, they cut proportional lengths on them.

In other words, the small triangle is a scaled-down copy of the big one: all its lengths are multiplied by the same number.

The two configurations to recognise

In both cases the setting is the same: two lines meeting at a point $A$ , and two parallel lines cutting them.

Nested triangles

Points $M$ and $N$ lie on the segments $[A B]$ and $[A C]$ : the small triangle $A M N$ sits inside the big triangle $A B C$ .

Nested triangles: M on [AB], N on [AC], and (MN) parallel to (BC).

The butterfly

Points $M$ and $N$ are on the other side of $A$ : triangles $A M N$ and $A B C$ are opposite at the vertex, like the two wings of a butterfly.

The butterfly: B, A, M collinear, C, A, N collinear, and (MN) parallel to (BC).

The condition to check first

Before writing a single fraction, you must be able to state two things:

points $A$ , $M$ , $B$ on one side, and $A$ , $N$ , $C$ on the other, are collinear;
lines $(M N)$ and $(B C)$ are parallel, because the question says so or because you have proved it.

Only then does the theorem apply:

$\frac{A M}{A B} = \frac{A N}{A C} = \frac{M N}{B C}$

You don’t start with the lengths: you start with the parallel lines.

A complete example

In triangle $A B C$ , $M$ lies on $[A B]$ and $N$ on $[A C]$ , with $(M N) ∥ (B C)$ . We know that $A M = 3$ cm, $A B = 5$ cm, $A N = 4.2$ cm and $M N = 2.4$ cm. Find $A C$ and $B C$ .

Points $A$ , $M$ , $B$ and $A$ , $N$ , $C$ are collinear, and $(M N) ∥ (B C)$ : Thales’ theorem applies.
Write the ratios starting from $A$ : $\frac{3}{5} = \frac{4.2}{A C} = \frac{2.4}{B C}$ .
Cross-multiply: $A C = \frac{5 \times 4.2}{3} = 7$ cm.
Likewise: $B C = \frac{5 \times 2.4}{3} = 4$ cm.

The right reflex: every ratio starts from the intersection point $A$ . On top, a length of the small triangle; underneath, the matching length of the big one, always in the same order.

Thales or its converse?

Thales’ theorem is for finding a length when you know two lines are parallel.

Its converse does the opposite: it proves that two lines are parallel when the ratios are equal, provided the points are collinear in the same order. And if the ratios differ, the contrapositive shows the lines are not parallel.

To choose, ask one question: what are you looking for? A length: the theorem. Parallel lines: the converse.

Two exercises, as in Coach

Here are two exercises as they appear in Coach, with an answer written line by line in the maths field.

Exercise 1

3 points~5 min

In a butterfly, lines $(B M)$ and $(C N)$ meet at $A$ , and $(M N) ∥ (B C)$ . We know that $A M = 2$ cm, $A B = 6$ cm and $B C = 9$ cm. Find $M N$ .

ABAM=BCMN62=9MNMN=62×9=3cm

Reproduction of an exercise in Coach, with a written answer (sample content).

Exercise 2

4 points~5 min

Points $A$ , $M$ , $B$ are collinear in that order, and so are $A$ , $N$ , $C$ . $A M = 4$ , $A B = 10$ , $A N = 6$ and $A C = 15$ . Are lines $(M N)$ and $(B C)$ parallel?

ABAM=104=0.4ACAN=156=0.4ABAM=ACAN, so (MN)∥(BC)

Reproduction of an exercise in Coach, with a written answer (sample content).

The checklist before calculating

Find the two intersecting lines and their intersection point.
Check the parallel lines: given in the question, or proved.
Write the three ratios starting from the intersection point, small triangle on top.
Keep the two useful ratios, then cross-multiply.
Conclude, with the unit.