Chapter 8.5 — Exercise 8.4 — Pythagoras Theorem

An Altitude Creates Two Similar Triangles

In a right triangle ABC, right-angled at B, drop a perpendicular BD from the right angle down to the hypotenuse AC. That one altitude quietly creates two smaller right triangles, and both turn out similar to the original — and to each other.

△ADB~△ABC by AA: they share ∠A, and both ∠ADB and ∠ABC are 90°. The same reasoning gives △BDC~△ABC using the shared ∠C. Since both smaller triangles are similar to the same original triangle, they're automatically similar to each other too. Every triangle in this figure — the original and both smaller pieces — shares the identical set of three angle measures; only their sizes differ, which is precisely what makes every side-length ratio between them trustworthy.

Deriving Pythagoras From These Two Similarities

Each of the two similarities above hands over its own proportion, and both proportions turn into a squared side length once cross-multiplied:

Adding these two results together, the right-hand side becomes (AD+CD)·AC — and since AD+CD is simply the whole hypotenuse AC, that product becomes AC·AC = AC².

The square of the hypotenuse equals the sum of the squares of the other two sides — reached without ever measuring an angle beyond the original right angle, purely from two similar-triangle proportions added together. The AD+CD=AC step is the quiet hinge the whole proof turns on: it's only true because D sits somewhere on the hypotenuse itself, splitting it into exactly two pieces that add back up to the whole — which is precisely why the altitude has to be dropped onto the hypotenuse, not onto either of the two shorter legs, for this particular proof to work.

The Converse, Built the Other Way

If a triangle's longest side squared equals the sum of the other two sides squared, that triangle must be right-angled — proved by building a genuinely right-angled triangle with the same two shorter sides, showing its hypotenuse also matches the given triangle's longest side by the Pythagoras theorem itself, and concluding the two triangles are congruent (SSS). Congruent triangles share every angle, so the given triangle's angle at that vertex must be 90° too. This converse is what actually makes the theorem useful for a completely different kind of question than the forward direction answers: instead of finding an unknown side of an already-known right triangle, it tests whether a triangle is right-angled in the first place, using nothing but its three side lengths.

Two Real-World Applications

• A guy-wire and a pole: an 18 m pole is held by a 24 m wire staked into the ground. How far from the pole's base should the stake sit? PR² = 24²−18² = 576−324 = 252 = 36×7, so PR = 6√7 m (≈15.87 m) — a genuinely irrational distance, not a round number, so leaving the answer as 6√7 rather than a rounded decimal keeps it exact.
• Two poles, one gap: poles of height 6 m and 11 m stand 12 m apart. The distance between their tops is the hypotenuse of a right triangle with legs 12 m (the horizontal gap) and 5 m (the height difference, 11−6): 12²+5² = 144+25 = 169, so the distance is 13 m — a clean result, since 5-12-13 is one of the well-known Pythagorean triples.

Both problems share the same first move, worth naming explicitly: turn the real-world setup — a wire, a pair of poles — into a right triangle first, identifying which measurement plays hypotenuse and which two play the legs, before writing down a single equation. The wire problem gives the hypotenuse and one leg directly; the two-poles problem gives neither directly, and instead needs a short horizontal-and-vertical-difference construction before the legs are even identified.

Two Short Structural Proofs

• A geometric mean, hidden in the altitude: in right triangle PQR (right-angled at P), altitude PM to the hypotenuse creates △RMP~△PMQ, giving PM/QM = MR/PM — rearranged, PM² = QM·MR. The altitude's length is the geometric mean of the two pieces it splits the hypotenuse into.
• An isosceles right triangle: if △ABC is right-angled at C with AC=BC, then AB²=AC²+BC²=AC²+AC², giving AB² = 2AC² directly — no similar triangles needed at all here, just the theorem applied to two already-equal legs.

The geometric-mean result is worth a second look for how it repackages the same altitude construction from earlier in this exercise: PM is genuinely the altitude to the hypotenuse in a right triangle, so it obeys the identical similar-triangle relationships already derived above — this problem is really just extracting one specific consequence, PM²=QM·MR, that was already implicit in the general derivation the whole exercise opened with.

Shapes Other Than Squares, Built on the Same Sides

Pythagoras' theorem is often pictured with squares built on each side, but the same area relationship holds for any similar shape built on all three sides — including equilateral triangles.

The equilateral triangle built on the hypotenuse always has the same area as the two smaller equilateral triangles on the legs combined — the identical relationship as the classic squares version, just rebuilt with a different shape entirely.

A related result connects a square to its own diagonal: for a square of side a, the diagonal measures √2·a. The equilateral triangle built on that diagonal has area (√3/4)(√2a)² = (√3/2)a², while the equilateral triangle built on the square's plain side has area (√3/4)a² — exactly half as much. The equilateral triangle on a square's side is always half the area of the one built on its diagonal. The √2 factor here comes from exactly the same place it always does — a diagonal splitting a square into two right-angled isosceles triangles, the same shape this exercise's isosceles-right-triangle result already covered above.

The Chapter, Complete

Similar Triangles built up from a single theorem to a complete toolkit: the introduction established BPT and its converse, Exercise 8.1 proved segments parallel and triangles isosceles from it, Exercise 8.2 formalised AA/SSS/SAS similarity, Exercise 8.3 connected side ratios to area ratios, and this exercise used the same similar-triangle machinery to derive Pythagoras' theorem itself from scratch. The next chapter, Tangents and Secants to a Circle, returns to right angles in a new setting — the angle a tangent always makes with a circle's radius at the point of contact.