Chapter 4.1, Problem 22MS

A Pythagorean Theorem for triangles other than right triangles. Suppose we have a triangle that is not a right triangle. For example, consider:

Now, if we drop a perpendicular line from the top vertex down to the side of length a and we cut that side into two pieces of lengths, $a\text{'}anda",$ we would have:

It turns out that $a^2+b^2=c^2+2aa\text{'}.$

Use the Pythagorean Theorem with the two right triangles in the preceding picture to produce two equations. Subtract one equation from the other and notice that the $h^2$ terms drop out. Now, deduce the preceding formula (remember that $a=a\text{'}+a^{"},$ so you can solve for co. Once you have proved this formula, show what happens if the angle between sides a and b is 90^o. Notice that you have actually proved the “converse” of the Pythagorean Theorem: If a triangle has sides with lengths a, b, and c satisfying $a^2+b^2=c^2,$ then it is a right triangle.