Proving the Converse of the Pythagorean Theorem | Grade 8 Math | LevelKite

You already know a proof of the Pythagorean theorem. Here is the exact statement this lesson builds on.

A right triangle has one right angle, which is an angle of $90^\circ$ . The two sides that form the right angle are the legs. The side opposite the right angle is the hypotenuse. It is the one side that does not touch the right angle's corner. The theorem says: if a triangle is a right triangle with legs of lengths $a$ and $b$ and a hypotenuse of length $c$ , then $a^2 + b^2 = c^2$ .

The converse of an if-then statement swaps its two parts. The theorem starts with a right angle and ends with the equation. Its converse starts with the equation and ends with a right angle.

Key concept

The converse and why it is true

Suppose a triangle has side lengths $a$ , $b$ and $c$ , all positive and in the same unit, with $c$ the largest. If $a^2 + b^2 = c^2$ , then the triangle is a right triangle, and its right angle is the angle opposite side $c$ .

Why: build a right triangle with legs $a$ and $b$ . The Pythagorean theorem forces its hypotenuse to have length $c$ . The two triangles then have the same three side lengths, so they are congruent. That means slides, turns and flips can move one exactly onto the other. So the angle opposite $c$ in your triangle equals the built right angle.

The proof uses two facts about triangles.

You can build a right triangle with any two legs you choose. Draw a right angle. Starting from its corner, mark one positive length on each of its sides. Then join the two marks.
Side-side-side (SSS) congruence. Suppose the three side lengths of one triangle equal the three side lengths of another, pair by pair. Then the triangles are congruent. This makes sense, because a triangle built from three stiff sticks cannot change its shape. In congruent triangles, the matching angles are equal too. These are the angles that sit between matching pairs of sides.

Here is the plan carried out on one triangle, drawn exactly. In the figure, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. One grid unit is the side of one grid square. Triangle $ABC$ is the triangle you are testing. Side $AB$ runs from Point A at $(1, 1)$ to Point B at $(6, 1)$ , so $AB = 6 - 1 = 5$ . Point C is at $(4.2, 3.4)$ , placed so that $CA = 4$ and $CB = 3$ exactly. You cannot count slanted lengths on the grid, so the example gives them to you. Triangle $DEF$ is the right triangle you will build.

Triangle

ABC

has sides

3

,

4

and

5

. The built right triangle

DEF

has legs

FE = 3

and

FD = 4

along grid lines.

Worked example

Sides 3, 4 and 5, drawn exactly

Triangle $ABC$ in the figure has sides $BC = 3$ , $CA = 4$ and $AB = 5$ grid units. Use the plan of the proof to explain why angle $C$ , the angle at Point C, is a right angle.

Check the conditions of the converse. All three lengths are positive. The largest is AB = 5, so c = 5, a = BC = 3 and b = CA = 4. Square the two shorter sides and add them. Then square the longest side. Both results are 25, so the equation holds.
$3^2 + 4^2 = 9 + 16 = 25, \qquad 5^2 = 25$
Build triangle DEF with a right angle at F, leg EF = 3 and leg FD = 4. In the figure, Point F is at (8, 1), Point E is at (11, 1) and Point D is at (8, 5). FE is horizontal, so its length is the change in x. FD is vertical, so its length is the change in y. A horizontal side and a vertical side meet at a right angle, so angle F is exactly 90 degrees.
$EF = 11 - 8 = 3, \qquad FD = 5 - 1 = 4$
Triangle DEF is a right triangle, and DE is opposite its right angle. So the Pythagorean theorem applies to DEF. Notice that you use the theorem on DEF, which you built to be right. You do not use it on ABC.
$DE^2 = 3^2 + 4^2 = 9 + 16 = 25$
Two numbers square to 25: 5 and negative 5. DE is a length, so it cannot be negative.
$DE = 5$
Pair up the sides of the two triangles. All three pairs are equal, so triangles ABC and DEF are congruent by SSS.
$BC = EF = 3, \qquad CA = FD = 4, \qquad AB = DE = 5$
Angle C lies between the sides of lengths 3 and 4, and so does angle F. That makes them matching angles. Matching angles of congruent triangles are equal.
$\text{angle } C = \text{angle } F = 90^\circ$

Answer

Angle $C$ is a right angle, so triangle $ABC$ is a right triangle whose right angle is opposite its longest side, $AB = 5$ .

Notice what the figure can show and what it cannot. As step 2 explained, the coordinates prove the right angle at Point F. The grid does not show the right angle at Point C directly, because both sides through Point C are slanted. Angle $C$ may look like $90^\circ$ , but looks and even protractor readings are only approximate. The steps of the example are what prove it. Still, one exact drawing shows only one triangle. To prove the converse for every triangle that meets the conditions, you write the same steps with letters, as in the table below.

The proof for every triangle that meets the conditions

Statement	Reason
Triangle $ABC$ has positive side lengths $BC = a$ , $CA = b$ and $AB = c$ , with $c$ largest and $a^2 + b^2 = c^2$ .	Given: these are the conditions of the converse.
Build triangle $DEF$ with a right angle at $F$ , $EF = a$ and $FD = b$ .	Any two positive lengths can be the legs of a right triangle.
Let $x$ be the length $DE$ . Then $a^2 + b^2 = x^2$ .	Pythagorean theorem: $DEF$ is a right triangle, and $DE$ is opposite its right angle.
$x^2 = c^2$	Both $x^2$ and $c^2$ equal the same number, $a^2 + b^2$ .
$x = c$	The numbers whose square is $c^2$ are $c$ and $-c$ , and a length cannot be negative.
Triangles $ABC$ and $DEF$ are congruent.	SSS: $BC = EF = a$ , $CA = FD = b$ and $AB = DE = c$ .
Angle $C$ equals angle $F$ , which is $90^\circ$ , so triangle $ABC$ is a right triangle.	Matching angles of congruent triangles are equal; $C$ and $F$ each lie between the sides of lengths $a$ and $b$ .

Worked example

Decimal sides listed out of order

A triangle has sides of $2.5$ cm, $1.5$ cm and $2$ cm, where cm means centimeters. Use the proof to explain why it is a right triangle, and say where its right angle is.

Name the sides. All three are positive and in centimeters. The largest is 2.5, so c = 2.5. The two shorter sides are a = 1.5 and b = 2.
$1.5 < 2 < 2.5$
Check the equation of the converse, writing out each square.
$1.5^2 + 2^2 = 2.25 + 4 = 6.25, \qquad 2.5^2 = 6.25$
Build a right triangle with legs 1.5 cm and 2 cm, and call its hypotenuse x. The Pythagorean theorem applies because this triangle is right and x is opposite its right angle. A length cannot be negative, so x is the positive number whose square is 6.25.
$x^2 = 1.5^2 + 2^2 = 6.25, \text{ so } x = 2.5$
Both triangles have sides of 1.5 cm, 2 cm and 2.5 cm, so they are congruent by SSS. The built right angle lies between the 1.5 cm and 2 cm legs. So the matching angle of the given triangle is also 90 degrees.

Answer

The triangle is a right triangle. Its right angle is between the $1.5$ cm and $2$ cm sides, opposite the $2.5$ cm side.

Worked example

When the equation fails

A triangle has sides of $4$ cm, $5$ cm and $6$ cm. Try the steps of the proof. Where do they stop, and does the triangle have a right angle?

All three lengths are positive. The largest is 6, so c = 6, a = 4 and b = 5. Check the equation of the converse. The two results differ, so the converse's condition is not met.
$4^2 + 5^2 = 16 + 25 = 41, \qquad 6^2 = 36$
Build a right triangle with legs 4 cm and 5 cm anyway, and call its hypotenuse x. The Pythagorean theorem applies to this built right triangle.
$x^2 = 4^2 + 5^2 = 41$
If x were 6, its square would be 36, not 41. So the built triangle has sides 4, 5 and x, which do not match 4, 5 and 6. You cannot use SSS, so the proof stops without showing a right angle.
$x^2 = 41 \neq 36 = 6^2, \text{ so } x \neq 6$
Could the triangle still have a right angle? If so, the theorem would apply, with the hypotenuse opposite that angle. The hypotenuse's square equals one leg's square plus the other leg's positive square, so the hypotenuse is longer than each leg. It would be the 6 cm side, and 16 + 25 would equal 36. It does not, so the triangle has no right angle.

Answer

The proof stops at the step that needs $x = 6$ , because $4^2 + 5^2 = 41$ but $6^2 = 36$ . Reasoning with the Pythagorean theorem then shows that the triangle has no right angle.

Takeaway: take a triangle with positive side lengths $a$ , $b$ and $c$ , with $c$ the largest. If $a^2 + b^2 = c^2$ , the angle opposite $c$ is a right angle. To prove it, build a right triangle with legs $a$ and $b$ . The Pythagorean theorem forces its hypotenuse to equal $c$ , and SSS makes the two triangles congruent. So the angle opposite $c$ matches the built right angle. A drawing can show one case, but only the argument with letters proves the converse for every triangle that meets the conditions.