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# 1.2: Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12

The Pythagorean Theorem allows you to find the lengths of the sides of a right triangle, which is a triangle with one 90\begin{align*}90^\circ\end{align*} angle (known as the right angle). An example of a right triangle is depicted below.

A right triangle is composed of three sides: two legs, which are labeled in the diagram as leg1\begin{align*}leg_{1}\end{align*} and leg2\begin{align*}leg_{2}\end{align*}, and a hypotenuse, which is the side opposite to the right angle. The hypotenuse is always the longest of the three sides. Typically, we denote the right angle with a small square, as shown above, but this is not required.

The Pythagorean Theorem states that the length of the hypotenuse squared equals the sum of the squares of the two legs. This is written mathematically as:

(leg1)2+(leg2)2=(hypotenuse)2\begin{align*}({\color{green}leg}_{\color{green}1})^2 + ({\color{blue}leg}_{\color{blue}2})^2 = ({\color{red}hypotenuse})^2 \end{align*}

To verify this statement, first explicitly expressed by Pythagoreans so many years ago, let’s look at an example.

### Example 1

Consider the right triangle below. Does the Pythagorean Theorem hold for this triangle?

Solution

As labeled, this right triangle has sides with lengths 3, 4, and 5. The side with length 5, the longest side, is the hypotenuse because it is opposite to the right angle. Let’s say the side of length 4 is leg1\begin{align*}leg_{1}\end{align*} and the side of length 3 is leg2\begin{align*}leg_{2}\end{align*}.

Recall that the Pythagorean Theorem states:

(leg1)2+(leg2)2=(hypothenuse)2\begin{align*}(leg_1)^2 + (leg_2)^2 = (hypothenuse)^2\end{align*}

If we plug the values for the side lengths of this right triangle into the mathematical expression of the Pythagorean Theorem, we can verify that the theorem holds:

(4)2+(3)216+925=(5)2=25=25\begin{align*}(4)^2+(3)^2 &=(5)^2 \\ 16+9 &=25 \\ 25 &=25\end{align*}

Although it is clear that the theorem holds for this specific triangle, we have not yet proved that the theorem will hold for all right triangles. A simple proof, however, will demonstrate that the Pythagorean Theorem is universally valid.

### Proof Based on Similar Triangles

The diagram below depicts a large right triangle (triangle ABC\begin{align*}{\color{red}ABC}\end{align*}) with an altitude (labeled h\begin{align*}h\end{align*}) drawn from one of its vertices. An altitude is a line drawn from a vertex to the side opposite it, intersecting the side perpendicularly and forming a 90\begin{align*}90^\circ\end{align*} angle.

In this example, the altitude hits side AB\begin{align*}AB\end{align*} at point D\begin{align*}D\end{align*} and creates two smaller right triangles within the larger right triangle. In this case, triangle ABC\begin{align*}{\color{red}ABC}\end{align*} is similar to triangles CBD\begin{align*}{\color{green}CBD}\end{align*} and ACD\begin{align*}{\color{blue}ACD}\end{align*}. When a triangle is similar to another triangle, corresponding sides are proportional in lengths and corresponding angles are equal. In other words, in a set of similar triangles, one triangle is simply an enlarged version of the other.

Similar triangles are often used in proving the Pythagorean Theorem, as they will be in this proof. In this proof, we will first compare similar triangles ABC\begin{align*}{\color{red}ABC}\end{align*} and CBD\begin{align*}{\color{green}CBD}\end{align*}, then triangles ABC\begin{align*}{\color{red}ABC}\end{align*} and ACD\begin{align*}{\color{blue}ACD}\end{align*}.

Comparing Triangles ABC\begin{align*}{\color{red}ABC}\end{align*} and CBD\begin{align*}{\color{green}CBD}\end{align*}

In the diagram above, side AB\begin{align*}{\color{red}AB}\end{align*} corresponds to side CB\begin{align*}{\color{green}CB}\end{align*}. Similarly, side BC\begin{align*}{\color{red}BC}\end{align*} corresponds to side BD\begin{align*}{\color{green}BD}\end{align*}, and side CA\begin{align*}{\color{red}CA}\end{align*} corresponds to side DC\begin{align*}{\color{green}DC}\end{align*}. It is possible to tell which side corresponds to the appropriate side on a similar triangle by using angles; for example, corresponding sides AB\begin{align*}{\color{red}AB}\end{align*} and CB\begin{align*}{\color{green}CB}\end{align*} are both opposite a right angle.

Because corresponding sides are proportional and have the same ratio, we can set the ratios of their lengths equal to one another. For example, the ratio of side AB\begin{align*}{\color{red}AB}\end{align*} to side BC\begin{align*}{\color{red}BC}\end{align*} in triangle ABC\begin{align*}{\color{red}ABC}\end{align*} is equal to the ratio of side CB\begin{align*}{\color{green}CB}\end{align*} to corresponding side BD\begin{align*}{\color{green}BD}\end{align*} in triangle CBD\begin{align*}{\color{green}CBD}\end{align*}:

length of ABlength of BC=length of CBlength of BD\begin{align*}\frac{\text{length of } {\color{red}AB}}{\text{length of } {\color{red}BC}} = \frac{\text{length of } {\color{green}CB}}{\text{length of } {\color{green}BD}}\end{align*}

Written with variables, this becomes:

ca=ax\begin{align*}\frac{c}{a} = \frac{a}{x}\end{align*}

Next, we can simplify this equation by multiplying both sides of the equation by a and x:

x×a×ca=ax×x×a\begin{align*}{x} \times {a} \times \frac{c}{a} = \frac{a}{x} \times {x} \times {a}\end{align*}

With simplification, we obtain:

cx=a2\begin{align*}{cx} &=a^2\end{align*}

Comparing Triangles ABC\begin{align*}{\color{red}ABC}\end{align*} and ACD\begin{align*}{\color{blue}ACD}\end{align*}

Triangle ABC\begin{align*}{\color{red}ABC}\end{align*} is also similar to triangle ACD\begin{align*}{\color{blue}ACD}\end{align*}. Side AB\begin{align*}{\color{red}AB}\end{align*} corresponds to side CA\begin{align*}{\color{blue}CA}\end{align*}, side BC\begin{align*}{\color{red}BC}\end{align*} corresponds to side CD\begin{align*}{\color{blue}CD}\end{align*}, and side AC\begin{align*}{\color{red}AC}\end{align*} corresponds to side DA\begin{align*}{\color{blue}DA}\end{align*}.

Using this set of similar triangles, we can say that:

\begin{align*}\frac{\text{length of } {\color{blue}CA}}{\text{length of } {\color{blue}DA}} = \frac{\text{length of } {\color{red}AB}}{\text{length of } {\color{red}AC}}\end{align*}

Written with variables, this becomes:

\begin{align*}\frac{b}{c-x} = \frac{c}{b}\end{align*}

Similar to before, we can multiply both sides of the equation by c-x and b:

\begin{align*} {b} \times {(c-x)} \times \frac{b}{c-x} &= \frac{c}{b} \times {b} \times {(c-x)} \\ b^2 &= c(c-x) \\ c^2 &= cx + b^2\end{align*}

Earlier, we found that \begin{align*}cx = a^2\end{align*}. If we replace \begin{align*}cx\end{align*} with \begin{align*}a^2\end{align*}, we obtain \begin{align*}c^2 = a^2 + b^2\end{align*}. This is just another way to express the Pythagorean Theorem. In the triangle \begin{align*}{\color{red}ABC}\end{align*}, side \begin{align*}c\end{align*} is the hypotenuse, while sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the two legs of the triangle.

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