Pythagorean theorem — proof (by similar triangles, rearrangement), converse

Overview

The Pythagorean theorem is the cornerstone of Euclidean geometry, relating the three sides of any right triangle. It states that in a right triangle, the square of the hypotenuse (the longest side opposite the right angle) equals the sum of squares of the other two sides.

Theorem

Why this matters: This theorem bridges algebra and geometry. It lets us compute distances, proves the distance formula, underlies trigonometry, and appears in vector analysis, complex numbers, and even special relativity.

Proof 1: Similar Triangles (Algebraic)

Setup: Start with right triangle $ABC$ with right angle at $C$. Drop a perpendicular from $C$ to the hypotenuse $AB$, meeting it at point $D$.

Why drop the perpendicular? It creates three similar triangles: the original $\triangle ABC$ and two smaller triangles $\triangle ACD$ and $\triangle CBD$. Similar triangles have proportional sides—we'll exploit that.

Step-by-step derivation

Let's denote:

• Leg $BC = a$, leg $AC = b$, hypotenuse $AB = c$
• Perpendicular $CD = h$
• Segments $AD = x$ and $DB = y$, so $x + y = c$

Step 1: Identify the similar triangles.

All three triangles have the same angles:

• $\triangle ABC \sim \triangle ACD \sim \triangle CBD$

Why are they similar? Each has a right angle, and they share acute angles. AA similarity criterion.

Step 2: Write proportionality relationships.

For $\triangle ABC \sim \triangle ACD$: $\frac{AC}{AB} = \frac{AD}{AC}$

Why this ratio? We're matching corresponding sides. $AC$ corresponds to the hypotenuse of the smaller triangle $\triangle ACD$.

Substituting lengths: $\frac{b}{c} = \frac{x}{b}$

Cross-multiply: $b^2 = cx \quad \text{..(1)}$

Step 3: Apply similarity to the other pair.

For $\triangle ABC \sim \triangle CBD$: $\frac{BC}{AB} = \frac{BD}{BC}$

$\frac{a}{c} = \frac{y}{a}$

Cross-multiply: $a^2 = cy \quad \text{...(2)}$

Step 4: Add equations (1) and (2).

$a^2 + b^2 = cy + cx = c(y + x)$

Why add them? We want $a^2 + b^2$ one side. Notice that $x + y = c$ (the two segments make up the hypotenuse).

$a^2 + b^2 = c \cdot c = c^2$

∴ Proven! $\boxed{a^2 + b^2 = c^2}$

Proof 2: Rearangement (Geometric)

Big idea: Arrange four copies of the same right triangle around a square, then compare areas two different ways.

Step 1: Outer square construction.

Take four identical right triangles with legs $a, b$ and hypotenuse $c$. Arrange them around the perimeter of a square so:

• The hypotenuses form the outer edges
• The legs create a tilted inner square

The outer square has side length $a + b$ (one leg plus another leg).

Step 2: Calculate total area—Method 1.

Outer square area: $(a+b)^2 = a^2 + 2ab + b^2$ $\text{Total area} = 4 \cdot \frac{1}{2}ab + c^2 = 2ab + c^2$ $4 \times \frac{1}{2}ab + c^2 = 2ab + c^2$

Step 4: Equate the two expressions.

Since both represent the same total area: $a^2 + 2ab + b^2 = 2ab + c^2$

Subtract $2ab$ from both sides: $a^2 + b^2 = c^2$

∴ Proven! This visual proof shows theorem is about area conservation.

The Converse

Why is the converse important? It's a test for right angles. You can verify a right angle by measuring sides, without needing a protractor.

Proof of the Converse

Given: Triangle with sides $a, b, c$ where $c^2 = a^2 + b^2$.

To prove: The angle opposite $c$ is a right angle.

Strategy: Construct a known right triangle and prove our triangle is congruent to it.

Step 1: Construct a reference right triangle.

Build a right triangle with legs $a$ and $b$. By the Pythagorean theorem (forward direction), its hypotenuse has length $\sqrt{a^2 + b^2}$.

Step 2: Compare hypotenuses.

Given: $c^2 = a^2 + b^2$, so $c = \sqrt{a^2 + b^2}$.

The reference triangle's hypotenuse also equals $\sqrt{a^2 + b^2}$.

Therefore: Both triangles have all three sides equal ($a, b, c$).

Step 3: Apply SS congruence.

By SSS (Side-Side-Side) congruence, the two triangles are congruent.

Conclusion: Since the reference triangle has a right angle, and the triangles are congruent, our original triangle must also have a right angle. ∎

Common Mistakes

Key Formulas Summary

Connections

• Distance Formula in Coordinate Geometry — direct application of Pythagorean theorem
• Pythagorean Triples — integer solutions like3-4-5, 5-12-13
• Trigonometric Ratios — derived from right triangles, foundation for sine and cosine
• Law of Cosines — generalization to non-right triangles
• 3D Distance Formula — extends to $d = \sqrt{x^2 + y^2 + z^2}$
• Complex Numbers — magnitude $|z| = \sqrt{a^2 + b^2}$ for $z = a + bi$
• Vectors and Magnitude — $\|\vec{v}\| = \sqrt{v_x^2 + v_y^2}$

#flashcards/maths

What does the Pythagorean theorem state for a right triangle? :: For a right triangle with legs $a$ and $b$, and hypotenuse $c$: $c^2 = a^2 + b^2$

How do you test if a triangle with sides 8, 15, 17 is a right triangle? :: Check if $8^2 + 15^2 = 17^2$. Compute: $64 + 225 = 289 = 289$ ✓ So yes, it's a right triangle.

Concept Map

Hinglish (regional understanding)