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The Pythagorean theorem: three proofs

Geometry · level 6, lesson 1 of 6

A question to start

There is a book, The Pythagorean Proposition by Elisha Scott Loomis, that collects more than 350 different proofs of a single fact about right triangles. People have never stopped finding new ones. Euclid gave one; Albert Einstein found one as a schoolboy (so he later wrote); and James Garfield, a member of the United States Congress who later became president, published one in a teachers' magazine in 1876.

Why would anyone need 350 proofs of the same thing? Because each proof shows the fact from a different side, and some of those sides lead somewhere new. This lesson gives you three, each with a different idea inside it: moving pieces around, scaling, and adding up areas. By the end you will not just know that \(a^2 + b^2 = c^2\); you will be able to explain to anyone why it has to be true.

The idea

The statement

A right triangle has one right angle (90°). The side opposite the right angle is the hypotenuse, always the longest side. The two sides that make the right angle are the legs.

Pythagorean theorem. In a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the squares on the legs add up to the square on the hypotenuse:

\[a^2 + b^2 = c^2\]

"\(a^2\)" can be read literally as the area of a square with side \(a\). So the theorem says: build a square on each side of the triangle; the two squares on the legs, together, have exactly the same area as the square on the hypotenuse.

For the triangle with sides 3, 4 and 5: the squares have areas 9, 16 and 25, and \(9 + 16 = 25\).

It is named after Pythagoras, a Greek thinker of about 2500 years ago, but it was known long before him. A Babylonian clay tablet called Plimpton 322, more than a thousand years older, lists sets of numbers that fit it, and Chinese and Indian texts use it too. The story goes that the followers of Pythagoras were the first to prove it, but no proof from that time survives.

What a proof has to do

Checking 3, 4, 5 is not a proof. Neither is checking a hundred triangles. A proof must work for every right triangle at once, so it uses letters \(a\), \(b\), \(c\) for sides of any length and argues only from facts that are always true. Each of the three proofs below does that.

Proof 1: rearranging squares

Take four copies of the right triangle and a big square frame with side \(a + b\). Arrange the four triangles in the frame in two different ways:

c²ab a²b² abba

Both big squares have the same area, \((a + b)^2\). Both contain the same four orange triangles. Take the four triangles away from each, and what is left must have the same area too. On the left, what is left is the tilted square \(c^2\). On the right, it is the two squares \(a^2\) and \(b^2\). So

\[c^2 = a^2 + b^2\]

One check. Is the blue shape on the left really a square? Its four sides are all hypotenuses, so all equal to \(c\). At each of its corners, three angles sit along a side of the frame: the two acute angles of a triangle and the corner of the blue shape. The two acute angles of a right triangle add to \(90°\) (the three angles add to \(180°\) and one of them is \(90°\)). Angles on a straight line add to \(180°\), so the corner is \(180° - 90° = 90°\). Four equal sides and four right angles: a square.

This proof needs no algebra at all. It is a proof by moving pieces, and it is perhaps the clearest of the 350.

Proof 2: similar triangles

This proof uses the idea from level 5, lesson 4: shapes that are scaled copies of each other. Two triangles with the same three angles are similar: one is a dilation of the other (maybe turned or flipped), so the ratios of matching sides are equal. (Level 7 studies similarity in depth.)

Take the right triangle \(ABC\) with the right angle at \(C\). Drop a perpendicular from \(C\) to the hypotenuse, meeting it at \(D\). This line is called the altitude to the hypotenuse. It splits \(c\) into two pieces, \(AD\) and \(DB\).

ABCD ba

Now there are three right triangles: the big one \(ABC\), the small orange one \(ACD\), and the blue one \(CBD\).

Step 1: they all have the same angles. The orange triangle \(ACD\) and the big triangle \(ABC\) share the angle at \(A\), and both have a right angle. Their third angles are then equal too, because all three add to \(180°\). So \(ACD\) is similar to \(ABC\), with \(A \leftrightarrow A\), \(D \leftrightarrow C\) (the right angles) and \(C \leftrightarrow B\). In the same way, the blue triangle \(CBD\) shares the angle at \(B\) with the big one and is similar to it.

Step 2: compare matching sides. In the big triangle, the leg \(AC = b\) next to angle \(A\), divided by the hypotenuse \(AB = c\), gives \(\dfrac{b}{c}\). In the orange triangle, the side next to angle \(A\) is \(AD\) and the hypotenuse is \(AC = b\). Similar triangles have equal ratios, so

\[\frac{AD}{b} = \frac{b}{c}, \qquad \text{which gives} \qquad b^2 = c \times AD.\]

The same comparison in the blue triangle, at angle \(B\), gives

\[\frac{DB}{a} = \frac{a}{c}, \qquad \text{which gives} \qquad a^2 = c \times DB.\]

Step 3: add.

\[a^2 + b^2 = c \times DB + c \times AD = c \times (DB + AD) = c \times c = c^2.\]

This proof explains something proof 1 does not: the square \(c^2\) on the hypotenuse splits into two rectangles, one with area \(a^2\) and one with area \(b^2\), cut along the line of the altitude. It is essentially the proof Euclid gave, in Book I of the Elements.

Proof 3: Garfield's trapezoid

Garfield's idea was to use half of proof 1's picture. Put two copies of the triangle at the two ends of a straight base, one with leg \(b\) along the base and one with leg \(a\) along it, and join the tops. You get a trapezoid (a four-sided shape with one pair of parallel sides; here the two upright sides \(a\) and \(b\) are parallel).

abab cc90°

The middle triangle has two sides equal to \(c\), and the angle between them, at the bottom, is a right angle: it is \(180°\) minus the two acute angles of the triangle, which add to \(90°\) (the same check as in proof 1).

Now find the trapezoid's area in two ways.

As a trapezoid: the area of a trapezoid is the average of the parallel sides times the distance between them. Here that is \(\dfrac{a + b}{2} \times (a + b) = \dfrac{(a + b)^2}{2}\).

As three triangles: two orange ones of area \(\tfrac12 ab\) each, and the blue one of area \(\tfrac12 c \times c\).

\[\frac{(a + b)^2}{2} = \frac12 ab + \frac12 ab + \frac12 c^2\]

Multiply both sides by 2: \((a + b)^2 = 2ab + c^2\). Multiply out the left side, \((a + b)^2 = a^2 + 2ab + b^2\). So \(a^2 + 2ab + b^2 = 2ab + c^2\), and taking \(2ab\) from both sides,

\[a^2 + b^2 = c^2.\]

The converse: it works backwards too

The converse of a statement swaps the "if" and the "then". The converse here is: if a triangle's sides satisfy \(a^2 + b^2 = c^2\), then it has a right angle (opposite \(c\)). That is a different claim and needs its own proof. Here is a short one using congruence.

Suppose triangle \(T\) has sides \(a\), \(b\), \(c\) with \(a^2 + b^2 = c^2\). Build a second triangle with a right angle between legs \(a\) and \(b\). By the theorem, its hypotenuse \(h\) satisfies \(h^2 = a^2 + b^2 = c^2\), so \(h = c\). Now both triangles have sides \(a\), \(b\), \(c\): by SSS (level 5, lesson 5) they are congruent, so \(T\) also has a right angle.

More is true. Open the hinge between sides \(a\) and \(b\) wider than 90° and the third side grows; close it and the third side shrinks. So with \(c\) the longest side:

  • \(a^2 + b^2 = c^2\): right angle;
  • \(a^2 + b^2 > c^2\): the angle opposite \(c\) is acute (less than 90°);
  • \(a^2 + b^2 < c^2\): the angle opposite \(c\) is obtuse (more than 90°).

Worked examples

1. Squares you can count. For the 5, 12, 13 triangle: the squares on the legs have areas \(25\) and \(144\), and \(25 + 144 = 169 = 13^2\). In proof 1, the frame has side \(5 + 12 = 17\) and area \(289\); the four triangles have area \(4 \times \tfrac12 \times 5 \times 12 = 120\); and \(289 - 120 = 169\), the tilted square.

2. Using the altitude. In proof 2's picture, suppose \(AD = 9\) and \(DB = 16\), so \(c = 25\).

  1. \(b^2 = c \times AD = 25 \times 9 = 225\), so \(b = 15\).
  2. \(a^2 = c \times DB = 25 \times 16 = 400\), so \(a = 20\).
  3. Check: \(15^2 + 20^2 = 225 + 400 = 625 = 25^2\).
  4. A bonus: the orange and blue triangles are similar to each other too, and comparing them gives \(CD^2 = AD \times DB = 9 \times 16 = 144\), so the altitude is \(CD = 12\).

3. Garfield with numbers. For \(a = 3\), \(b = 4\), \(c = 5\): the trapezoid's area is \(\dfrac{(3 + 4)^2}{2} = 24.5\). The three triangles: \(\tfrac12 \times 3 \times 4 = 6\), another 6, and \(\tfrac12 \times 5 \times 5 = 12.5\). Total \(6 + 6 + 12.5 = 24.5\). The two counts agree, as the proof promises.

4. The converse. Is the triangle with sides 9, 40, 41 right-angled? \(9^2 + 40^2 = 81 + 1600 = 1681\), and \(41^2 = 1681\). Equal, so yes: the right angle is opposite the 41 side.

Try it

  1. Show that 20, 21, 29 is a right triangle.
  2. For legs 6 and 8, find the area of the big frame in proof 1, the area of the four triangles, and the area of the tilted square. What is \(c\)?
  3. A triangle has sides 6, 7 and 9. Is its largest angle right, acute or obtuse?
  4. In proof 2's picture, \(AD = 4\) and \(DB = 9\). Find \(a^2\), \(b^2\) and the altitude \(CD\). Check that \(a^2 + b^2 = c^2\).
  5. Check Garfield's two area counts for \(a = 5\), \(b = 12\).
  6. The squares on the two legs of a right triangle have areas 50 cm² and 94 cm². How long is the hypotenuse?
  7. Explain in your own words why, in proof 1, the two big squares must have the same area left over once the triangles are taken away.
  8. Harder: a triangle has sides 7, 8 and 12. Without drawing, decide whether it has an obtuse angle, and say which side it is opposite.
Answers
  1. \(20^2 + 21^2 = 400 + 441 = 841 = 29^2\). By the converse, it has a right angle opposite the 29 side.
  2. Frame side \(6 + 8 = 14\), area \(196\). Triangles: \(4 \times \tfrac12 \times 6 \times 8 = 96\). Tilted square: \(196 - 96 = 100\), so \(c = \sqrt{100} = 10\).
  3. Largest side 9: \(6^2 + 7^2 = 36 + 49 = 85\) and \(9^2 = 81\). Since \(85 > 81\), the largest angle is acute (so the triangle is acute).
  4. \(c = 4 + 9 = 13\). \(b^2 = 13 \times 4 = 52\), \(a^2 = 13 \times 9 = 117\), and \(52 + 117 = 169 = 13^2\). The altitude: \(CD^2 = 4 \times 9 = 36\), so \(CD = 6\).
  5. Trapezoid: \(\dfrac{(5 + 12)^2}{2} = \dfrac{289}{2} = 144.5\). Triangles: \(30 + 30 + \tfrac12 \times 13^2\), where \(c = 13\) since \(25 + 144 = 169\); that is \(30 + 30 + 84.5 = 144.5\). They agree.
  6. The square on the hypotenuse has area \(50 + 94 = 144\) cm², so the hypotenuse is \(\sqrt{144} = 12\) cm.
  7. Both big squares have the same side \(a + b\), so the same area. Both contain four identical triangles with the same total area. Equal minus equal is equal, so the leftover regions, \(c^2\) on one side and \(a^2 + b^2\) on the other, have the same area.
  8. Longest side 12: \(7^2 + 8^2 = 49 + 64 = 113\) and \(12^2 = 144\). Since \(113 < 144\), the angle opposite the 12 side is obtuse.

Common mistake

Using the theorem on a triangle with no right angle. \(a^2 + b^2 = c^2\) is true only for right triangles. For the 6, 7, 9 triangle, \(6^2 + 7^2 \ne 9^2\). If there is no right angle (or you cannot show there is one), the theorem does not apply; level 7's cosine law is the version for every triangle.

Picking the wrong side as \(c\). \(c\) is always the hypotenuse, the side opposite the right angle, and the longest. When testing the converse, put the longest side on its own: test \(7^2 + 24^2\) against \(25^2\), not \(7^2 + 25^2\) against \(24^2\).

Where it leads