6.3 The Converse of the Pythagorean Theorem
We have seen that the converse of a theorem is not always true. However, the converse of the Pythagorean Theorem is true. In this section we will explore the converse of the Pythagorean Theorem and some of the resulting consequences.
- Theorem 6.3:
The Converse of the Pythagorean Theorem
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, and the side that is the sum of the squares of the other two sides is the hypotenuse.
Given:
$\triangle ABC$ with $c^2=a^2+b^2$ Prove: $\triangle ABC$ is a right triangle with hyptenuse $c$. |
Proof: We start by constructing a right triangle, $\triangle EFG$, with legs $a$ and $b$ and right angle at $G$. If we can show $\triangle EFG\cong\triangle ABC$, then $\angle G\cong\angle C$ making $\triangle ABC$ a right triangle, with the right angle at $C$(and hence with hypotenuse $c$).
Since $\triangle EFG$ is a right triangle, the Pythagorean Theorem says that $g^2=a^2+b^2$. But we have been given that $c^2=a^2+b^2$. By substitution, we then know that $g=c$. Since all three corresponding sides of the two triangles are congruent, the SSS Congruence Postulate ensures us that $\triangle EFG\cong\triangle ABC$. The definition of congruent triangles then assures us that $\angle G\cong\angle C$, making $\triangle ABC$ a right triangle with the right angle at $C$.
One of the outcomes of the Pythagorean Theorem and its converse that the Pythagorean Brotherhood might have discussed are right triangles with sides whose lengths are whole numbers. The most famous is the triangle whose lengths measure 3–4–5.
- Definition:
Any three positive integers $a$, $b$, and $c$ that satisfy the equation
$a^2+b^2=c^2$ form a Pythagorean
Triple.
If each number in the triple 3–4–5 is doubled, we get the triple 6–8–10. A triangle whose sides are 6, 8, and 10 is similar to a triangle whose sides are 3, 4, and 5 because the corresponding sides of the two triangles are in proportion. Since the triangle whose sides are 3, 4, and 5 is a right triangle, the triangle whose sides are 6, 8, and 10 is also a right triangle. Therefore, a 6–8–10 is also a Pythagorean Triple. In the same way, triangles with sides 9, 12, and 15, of 18, 24, and 30, are right triangles. These triangles are all similar to the 3–4–5 triangle. Because of this similarity, we refer to the 3–4–5 family of right triangles. In the table on the next page we list some of the common Pythagorean triples.
3, 4, 5 |
5, 12, 13 |
8, 15, 17 |
7, 24, 25 |
6, 8, 10 |
10, 24, 26 |
16, 30, 34 |
14, 48, 50 |
9, 12, 15 |
15, 36, 39 |
||
12,
16, 20 |
The top row in this table is the family to which the others in that column belong. This collection of families goes on and on. It is infinite.
We now consider what happens in a triangle if there is no right angle. With this result, it is possible to determine whether a triangle is acute or obtuse.
- Theorem 6.4
Case 1:
In an obtuse triangle, the square of the longest side is greater than
the sum of the squares of the other two sides.
Case 2: In an acute triangle, the square of the longest side is less than the sum of the squares of the other two sides.
We consider these two situations simultaneously. Here we have two triangles, one obtuse and the other acute. In each case, let the length of the longest side be $c$. In each case, we drop the altitude from vertex $B$ to side $\overline{AC}$. In the obtuse triangle, this falls outside the triangle; in the acute triangle, it falls inside the triangle. In the figure below, drag point $B$ to change $\triangle ABC$ into an acute, right, and obtuse triangle. Note the relationship between $a^2+b^2$ and $c^2$.
We now use the Pythagorean Theorem in each case to compute $c^2$. We arrange our work with the obtuse case on the left and the acute case on the right.
\begin{align} c^2&=(b+x)^2+h^2\\ &=b^2+2bx+x^2+b^2\\ &=b^2+(x^2+h^2)+2bx\\ &=b^2+a^2+2bx \end{align} |
\begin{align} c^2&=(b-x)^2+h^2\\ &=b^2-2bx+x^2+b^2\\ &=b^2+(x^2+h^2)-2bx\\ &=b^2+a^2-2bx \end{align} |
On the left, $c^2=a^2+b^2+(\text{ something extra })$, so
$c^2>a^2+b^2$.
On the right, $c^2=a^2+b^2-(\text{ something extra })$, so $c^2<a^2+b^2$.
On the right, $c^2=a^2+b^2-(\text{ something extra })$, so $c^2<a^2+b^2$.
Along with the Pythagorean Theorem, this gives us three non-overlapping situations.
Acute Triangles: Square of longest side is less than the sum of squares of the other two sides
Right Triangles: Square of longest side is equal to the sum of squares of the other two sides
Obtuse Triangles: Square of longest side is greater than the sum of squares of other two sides
Since this exhausts all the possibilities, we can use the results the other way around.
Example: Consider a triangle with sides 5, 7, and 10. Is this triangle acute, right, or obtuse?
Solution: Since $10^2>5^2+7^2$, the triangle must be obtuse.