The intercept theorem

Two rays from a point, cut by a pair of parallel lines, split into matching ratios.

The last few lessons compared two separate, similar figures. This one finds the same ratio hiding inside a single pair of crossing lines.

Two rays from a point O. A and B sit on one ray, C and D on the other, with BD always drawn parallel to AC. A live reading gives all four lengths from O, and checks OA:OB against OC:OD.

Two rays from a point O. On one ray, A and B; on the other, C and a derived point D, placed so that BD is always parallel to AC. A live reading gives OA, OB, OC and OD, and checks that OA:OB always matches OC:OD. This needs JavaScript switched on.

Drag O, A, B or C. The two ratios always match, however far out you drag them.

D itself is never dragged directly — it moves only to keep BD parallel to AC, and the matching ratios follow from that constraint alone.

Two rays sharing a start point, cut by a pair of parallel lines, is the setup behind the intercept theorem: the ray segments nearer O are to the ray segments further out in the same ratio, on both rays at once.

Why the two ratios have to match

Triangle OAC and triangle OBD share the angle at O, and — because AC is parallel to BD — the angle at A matches the angle at B, and the angle at C matches the angle at D. These are corresponding angles on the parallel lines AC and BD, cut by the transversals OA and OC, the same fact used earlier to show a transversal crossing two parallel lines makes equal corresponding angles.

Two matching angles are enough for similarity, which was G7.2’s whole point. So triangle OAC and triangle OBD are similar, and similar triangles have every pair of corresponding sides in one constant ratio — which was G7.3’s whole point:

\frac{OA}{OB} = \frac{OC}{OD}

The intercept theorem is similarity and constant ratio, applied to a figure where the two “separate” triangles happen to share a vertex and a pair of parallel sides.

This is the shape that keeps appearing wherever one length needs to be found from another without measuring it directly: a shadow and a stick at different distances from a light, a ladder leaning at different heights against a wall, or scaling a technical drawing from a single measured rung. Whenever two lines cross at a point and get cut by something parallel, this ratio is already sitting there, whether or not anyone draws the rest of the triangle.

One to watch out for.

This is also the reverse of the constant-ratio lesson’s logic: there, similarity was already known and length followed. Here, only the parallel lines are known, and similarity is what has to be argued first, before the ratio can be trusted.

On a ray from O, A is 4 cm out and B is 10 cm out. On a second ray from O, C is 6 cm out, and AC is parallel to BD. How far out is D?

$\frac{OA}{OB} = \frac{OC}{OD} \Rightarrow \frac{4}{10} = \frac{6}{OD} \Rightarrow OD = 6 \times \frac{10}{4} = 15$ cm.

Two rays from O carry OA = 3, OB = 12, OC = 5. If AC is parallel to BD, what is OD?

$\frac{3}{12} = \frac{5}{OD} \Rightarrow OD = 5 \times \frac{12}{3} = 20$ .

Two rays from O carry OA = 6, OB = 15, OD = 20, with AC parallel to BD. What is OC this time — note you’re solving for a different segment than before.

$\frac{OA}{OB} = \frac{OC}{OD} \Rightarrow \frac{6}{15} = \frac{OC}{20} \Rightarrow OC = 20 \times \frac{6}{15} = 8$ .

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.