Chapter 8.1 — Introduction — Basic Proportionality Theorem

What Makes Two Figures Similar

Two polygons with the same number of sides are similar when their corresponding angles are equal and their corresponding sides are in the same ratio. Both conditions matter together — a rectangle and a parallelogram can share equal corresponding angles without similar sides, and a right triangle and an equilateral triangle never share equal angles at all, so neither pair counts as similar. Congruent figures are really just a special case of similar figures — identical shape and identical size, rather than identical shape scaled to a different size — which is why every pair of congruent triangles is automatically similar, but not every pair of similar triangles is congruent.

• Always similar, regardless of size: any two circles; any two squares; any two equilateral triangles — every regular polygon with the same number of sides is automatically similar to every other one of that same kind.
• Never similar, no matter how you resize them: a rectangle and a parallelogram (equal angles, but sides aren't proportional the same way); a right triangle and an equilateral triangle (angles don't match at all).

Circles deserve a small note of their own here: every circle is similar to every other circle regardless of radius, since a circle has no corners or sides to compare ratios between in the first place — but two circles are only congruent, not just similar, once their radii are actually equal.

Similarity Between Triangles Specifically

For triangles, similarity carries the same two-part definition: △ABC ~ △DEF means ∠A=∠D, ∠B=∠E, ∠C=∠F, and AB/DE = BC/EF = AC/DF = K for some constant K. That constant K is the scale factor between the two triangles — K<1 shrinks the shape, K=1 makes it an exact congruent copy, and K>1 enlarges it. The order the vertices are written in matters as much as the equality itself: writing △ABC ~ △DEF is a claim that A corresponds to D, B to E, and C to F specifically, not just that the two triangles happen to be similar in some unspecified pairing — get the vertex order wrong and the individual side ratios written down will be wrong even though the triangles genuinely are similar.

The Basic Proportionality Theorem

Also called Thales' Theorem, this is the single fact the rest of the chapter builds on: if a line is drawn parallel to one side of a triangle, cutting the other two sides at distinct points, it divides those two sides in the same ratio.

The proof leans entirely on areas rather than angles. Joining BE and CD, and dropping perpendiculars from D and E onto the opposite sides, gives ar(△ADE)/ar(△BDE) = AD/DB (both triangles share the same height from E) and ar(△ADE)/ar(△CDE) = AE/EC (both share the same height from D). Since △BDE and △CDE sit on the identical base DE between the same two parallel lines, they always have equal area — so both ratios above equal the same third ratio, ar(△ADE)/ar(△BDE), which forces AD/DB = AE/EC directly. It's worth noticing what this proof does not rely on: nothing about angle measures, nothing about the specific shape of the triangle, only the parallel-lines-equal-area fact and two ratios built from areas that share a common height. That's exactly why the result holds for every triangle and every choice of parallel line, not just some special-case configuration.

Adding 1 to both sides of that ratio gives a second, equally useful form of the same fact, built from whole-side lengths instead of split segments — genuinely useful whenever a problem states the length of a full side rather than the two pieces it's split into:

The Converse Works Just as Often

Flip the theorem around and it's still true: if a line divides two sides of a triangle in the same ratio, that line must be parallel to the third side. This converse gets used at least as often as the theorem itself, since it's frequently the faster way to prove two segments parallel — compute two ratios, confirm they match, and parallelism follows immediately without any angle-chasing at all. Two segments can look parallel on a rough sketch without actually being parallel, which is exactly the situation this converse is built to settle with certainty: the ratio comparison either confirms or rules out parallelism definitively, regardless of how convincing the sketch looks.

Three Quick Ratio Checks

In △PQR, with E on PQ and F on PR, whether EF ∥ QR comes down entirely to comparing ratios:

The third row uses the whole-side form of the ratio (PQ/PE compared with PR/PF) rather than the split-segment form used in the first two rows — both versions of the theorem test the exact same underlying parallelism, just built from different pairs of measured lengths. Whichever form a problem hands over — split segments or whole sides — converting to the other form first is never necessary; the alternate form derived above guarantees both comparisons agree, so it's fine to work directly with whichever lengths are actually given.

Finding a Missing Segment

Given DE ∥ BC in a triangle, with AE=1.8 cm, EC=5.4 cm, and DB=7.2 cm, the theorem finds AD directly without measuring anything else:

Notice that the full length of side AB was never needed here — only the segment ratio and the one known length DB. That's the real practical payoff of this theorem: it finds an unknown segment length from a proportion alone, without ever requiring every other measurement in the triangle to be known first.

Where This Theorem Leads

Every proof in Exercise 8.1 builds directly on this one ratio fact and its converse — proving triangles isosceles, proving lines parallel, and chaining the theorem through two triangles sharing a common vertex. Exercise 8.2 then formalises the AA, SSS, and SAS similarity criteria this theorem quietly assumes throughout, Exercise 8.3 connects similar triangles' side ratios to their area ratios, and Exercise 8.4 uses this same proportionality idea to derive the Pythagoras theorem itself. Nothing in any of those four exercises introduces a genuinely new starting idea — every proof is either this theorem applied directly, its converse applied directly, or both applied one after the other through a shared vertex or a shared parallel line, which is worth keeping in mind whenever a later proof looks unfamiliar at first glance.