The Timpson Lyceum · Geometry (2024)

Similar Triangles & Pythagorean Theorem – Lecture 12 (Geometry)

Overview

This lecture finishes the chapter on similar triangles by collecting the shortcuts that let us conclude two triangles are similar, then puts those shortcuts to work proving the Pythagorean Theorem. We start from the definition of similarity and the angle-angle criterion, add a proportional-side line theorem, and build up similarity versions of the familiar SAS and SSS congruence criteria. The payoff is a chain of results about right triangles: dropping an altitude to the hypotenuse creates two smaller triangles similar to the original, which yields the geometric mean and a clean proof of $a^2 + b^2 = c^2$. We close with the converse of the Pythagorean Theorem and the fact that an angle inscribed in a semicircle is always a right angle.

Key Concepts

Definition of similar triangles. Triangles $ABC$ and $DEF$ are similar when their corresponding angles are congruent and their corresponding sides are proportional: $\angle A \cong \angle D,\quad \angle B \cong \angle E,\quad \angle C \cong \angle F,\qquad \frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}.$

Angle-Angle (AA) similarity. If two angles of one triangle are congruent to two angles of another, the triangles are similar. Only two angles are needed: since the three angles of every triangle sum to a straight angle, the third pair is automatically congruent (supplements of congruent sums are congruent).

Proportional-line theorem. A line that divides two sides of a triangle proportionally is parallel to the third side. In triangle $ABC$ with $D$ on $AB$ and $E$ on $AC$, $\frac{AD}{AB} = \frac{AE}{AC} \;\Longrightarrow\; DE \parallel BC.$ This is the workhorse used to manufacture parallel lines (and therefore equal angles) in the later proofs.

Side-Angle-Side (SAS) similarity. If an angle of one triangle is congruent to an angle of another and the two sides forming that angle are proportional, the triangles are similar: $\angle A \cong \angle A',\quad \frac{A'B'}{AB} = \frac{A'C'}{AC} \;\Longrightarrow\; \triangle ABC \sim \triangle A'B'C'.$ The proportional sides must be the two that enclose the congruent angle.

Side-Side-Side (SSS) similarity. If all three pairs of corresponding sides are proportional, the triangles are similar — no angle information required: $\frac{A'B'}{AB} = \frac{A'C'}{AC} = \frac{B'C'}{BC} \;\Longrightarrow\; \triangle ABC \sim \triangle A'B'C'.$

Altitude to the hypotenuse. Dropping an altitude from the right angle of a right triangle to the hypotenuse splits it into two smaller triangles, each similar to the original (and to each other). All three triangles share the same angles.

Geometric (mean proportional). The geometric mean of two numbers $p$ and $q$ is the value $x$ with $p : x = x : q$, so $x = \sqrt{pq}$ — distinct from the arithmetic mean $\tfrac{p+q}{2}$. With the altitude $h$ splitting the hypotenuse into segments $x$ and $y$, $\frac{x}{h} = \frac{h}{y} \;\Longrightarrow\; h^2 = xy \;\Longrightarrow\; h = \sqrt{xy},$ i.e. the altitude is the geometric mean of the two hypotenuse segments.

Pythagorean Theorem. In a right triangle with legs $a$, $b$ and hypotenuse $c$, $a^2 + b^2 = c^2$. It is proved here from the similarity of the three triangles created by the altitude, not by an area argument.

Converse of the Pythagorean Theorem. If a triangle has sides $a$, $b$, $c$ with $a^2 + b^2 = c^2$, then the angle opposite side $c$ is a right angle. Proved by constructing a right triangle on legs $a$ and $b$; its hypotenuse is $\sqrt{a^2+b^2} = c$, so it is congruent (SSS) to the given triangle, forcing the matching angle to be right.

Angle in a semicircle. A triangle inscribed in a semicircle, with the diameter as one side and the third vertex on the arc, is always a right triangle, with the right angle at the vertex on the arc (opposite the diameter/hypotenuse).

Worked Example

Proving the Pythagorean Theorem with similar triangles. Drop the altitude $h$ from the right angle to the hypotenuse $c$, splitting it into segments $x$ (adjacent to leg $a$) and $y$ (adjacent to leg $b$), so $x + y = c$. Each small triangle is similar to the whole, giving "short leg over hypotenuse" proportions; solve each for a segment, then recombine.

Multiplying the third line through by $c$ clears the denominators and leaves the Pythagorean relation.

Summary

• Two triangles are similar when corresponding angles are congruent and corresponding sides are proportional; AA (two congruent angle pairs) already guarantees it.
• A line cutting two sides of a triangle proportionally is parallel to the third side — the key tool for generating equal angles in the similarity proofs.
• SAS similarity (one congruent angle between two proportional sides) and SSS similarity (all three sides proportional) mirror the congruence criteria.
• The altitude to the hypotenuse of a right triangle produces two triangles similar to the original; the altitude is the geometric mean $\sqrt{xy}$ of the hypotenuse segments.
• Those similar triangles give a clean proof that $a^2 + b^2 = c^2$, and its converse ($a^2+b^2=c^2 \Rightarrow$ right angle) follows by an SSS-congruence construction.
• An angle inscribed in a semicircle (subtending the diameter) is always a right angle.