【People's Education Press】Junior High School Math, Grade 8, Part B: Unlocking Zhao Shuang's String Diagram – A Clever Proof of the Pythagorean Theorem

Ancient Chinese mathematician Zhao Shuang pioneered the 'String Diagram' proof method while annotating the 'Zhou Bi Suan Jing'. This diagram avoids complex axiomatic derivations and instead uses a purely geometric area dissection technique—'proof by shape to verify numbers'—to perfectly blend geometric intuition with algebraic rigor. By preparing four congruent right triangles (with legs a, b and hypotenuse c), arranging them like a windmill, we naturally form a square gap of side length (b - a) at the center, while the outer boundary seamlessly forms a large square of side length c!

From Geometry to Algebra: Simplifying Complex Equivalences

The core formula of the Pythagorean Theorem reveals the equality between the squares of the three sides of a right triangle. Using Zhao Shuang's String Diagram, we can easily establish an area equation and definitively prove this theorem:

Step 1: Establish the Area Equation

Observe the assembled string diagram,the total area of the large squarecan be calculated in two ways:

Method 1: Directly calculate the large square (side length c), resulting in an area of $c^2$.

Method 2: Calculate the internal components separately—the combined area of the four right triangles plus the area of the small central square.

Step 2: Algebraic Expansion and Simplification

Based on Method 2, the algebraic expression is: $4 \times (\frac{1}{2}ab) + (b-a)^2$.

Expand the perfect square term: $2ab + (b^2 - 2ab + a^2)$.

Combine like terms, canceling out $2ab$ and $-2ab$, yielding the final result: $a^2 + b^2$.

Therefore, $a^2 + b^2 = c^2$ is proven!

Variant Model: Garfield's Trapezoidal Proof

Interestingly, in 1876, U.S. President James A. Garfield proposed a remarkably elegant proof using a trapezoid. He used just two congruent right triangles, offset them vertically, and connected their top vertices to form a right trapezoid. By equating the trapezoid's area, given by the formula $\frac{1}{2}(a+b)(a+b)$, with the sum of the areas of the three internal triangles—including one isosceles right triangle—he similarly derived $a^2 + b^2 = c^2$.

Inverse and Forward Applications of the Pythagorean Theorem in Real Life

In practical surveying and construction, the Pythagorean Theorem is a powerful tool for finding unknown distances. For example, if the side length of an equilateral triangular truss is $6$, engineers need not measure directly; they simply draw a height to split it into two right triangles. Using the formula $3^2 + \text{height}^2 = 6^2$, the height can be instantly calculated as $3\sqrt{3}$.

Similarly, if someone walks 80 meters east on flat ground, then turns and walks 60 meters, and finally walks 100 meters back to the starting point, the fact that $80^2 + 60^2 = 100^2$ perfectly satisfies the core formula (a scaled-up version of the classic 3-4-5 Pythagorean triple by a factor of 20) proves that their first turn must have formed a $90^\circ$ right angle! This is a brilliant real-world validation of the converse of the Pythagorean Theorem.

🎯 Core Principle: The Pythagorean Theorem

In a right triangle, the sum of the squares of the two legs (a, b) always equals the square of the hypotenuse (c). Whether calculating side lengths, finding the distance between coordinate points, or verifying a right angle, this formula is the cornerstone of geometry and algebra.

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

QUESTION 1

As shown in the figure, there are two points $A(5, 0)$ and $B(0, 4)$ in the Cartesian coordinate system. Find the straight-line distance between them.

$\sqrt{41}$

$41$

$9$

$3$

QUESTION 2

As shown in the figure, the side length of an equilateral triangle is $6$. Find: (1) the length of the height $AD$; (2) the area of this triangle.

Height $3\sqrt{3}$, Area $9\sqrt{3}$

Height $3\sqrt{3}$, Area $18\sqrt{3}$

Height $3$, Area $9$

Height $6$, Area $18$

QUESTION 3

For what real values of $x$ is the algebraic expression $\frac{1}{\sqrt{2x-1}}$ defined in the real number system?

$x > \frac{1}{2}$

$x \ge \frac{1}{2}$

$x \neq \frac{1}{2}$

$x < \frac{1}{2}$

QUESTION 4

If the three side lengths of a triangle, $a, b, c$, satisfy $a^2 = c^2 - b^2$, is the triangle formed by these segments a right triangle? Why?

Yes, because it satisfies the converse of the Pythagorean Theorem (not because two lines are parallel and alternate angles are equal)

No, if two real numbers are equal, then their absolute values are equal

No, corresponding angles of congruent triangles are equal

No, within an angle, points equidistant from the two sides lie on the angle bisector

QUESTION 5

Xiaoming walks 80 meters east, then walks 60 meters in another direction, and finally walks 100 meters in a third direction to return exactly to his starting point. After walking 80 meters east, in which direction did Xiaoming walk?

South or North

West or East

Southeast or Northeast

Southwest or Northwest

Challenge: Garfield's Trapezoidal Puzzle

Applying Algebraic Identities and Geometric Dissection

In 1876, U.S. President James A. Garfield published a remarkably ingenious proof of the Pythagorean Theorem. Observe the geometric figure below: he used two identical right triangles, offset them vertically, and connected their top vertices to form a complete 'right trapezoid'.

Within this right trapezoid, what shape is the yellow region? What are its base and height?

Detailed Analysis:
The yellow triangle is aright isosceles triangle.
Since the bottom blue triangle and the upper-left blue triangle are congruent (both having legs a and b), let the angles at the base be $\alpha$ and $\beta$. We know that in a right triangle, $\alpha + \beta = 90^\circ$.
A straight angle (a straight line) is $180^\circ$. Subtracting these two acute angles leaves an angle of $180^\circ - 90^\circ = 90^\circ$ in the middle. Therefore, the yellow triangle is not only isosceles (with two sides being the hypotenuses c), but also a right triangle. Its base and height are both $c$.

Using the trapezoid area formula $S = \frac{1}{2}(\text{top base} + \text{bottom base}) \times \text{height}$, write the algebraic expression for the total area of the entire figure, then expand and simplify it.

Detailed Analysis:
Observe the entire large figure; it is a right trapezoid:

Top Base: Short side $b$
Bottom Base: Long side $a$
Height: The vertical height of the trapezoid is the sum of two segments, i.e., $(a + b)$

Substitute into the trapezoid area formula:
$S_{\text{total}} = \frac{1}{2}(a + b)(a + b) = \frac{1}{2}(a + b)^2$
Using the perfect square formula to expand:
$S_{\text{total}} = \frac{1}{2}(a^2 + 2ab + b^2) = \frac{1}{2}a^2 + ab + \frac{1}{2}b^2$.

Now, represent the total area again by calculating the sum of the areas of the three internal small triangles, and set up an equation with the result from Q2. How does this prove the Pythagorean Theorem?

Detailed Analysis:
The area of the trapezoid can also be viewed as the sum of the areas of the three internal triangles:
$S_{\text{total}} = 2 \times (\text{blue right triangle}) + 1 \times (\text{yellow isosceles right triangle})$
$S_{\text{total}} = 2 \times (\frac{1}{2}ab) + \frac{1}{2}c^2 = ab + \frac{1}{2}c^2$

Since both area calculation methods describe the same trapezoid, the two expressions must be equal:
$\frac{1}{2}a^2 + ab + \frac{1}{2}b^2 = ab + \frac{1}{2}c^2$
Subtract $ab$ from both sides of the equation:
$\frac{1}{2}a^2 + \frac{1}{2}b^2 = \frac{1}{2}c^2$
Finally, multiply both sides of the equation by 2:
$a^2 + b^2 = c^2$
President Garfield completed the proof of the Pythagorean Theorem in just two steps of area equivalence substitution!