Prove that if a right triangle has all sides of integral length, then it has at most one angle bisector of integral length.

One may prove by the law of cosines that the length of the angle bisector from C in triangle ABC with side lengths a, b, c, is given by:

x²=ab[1 - c²/(a+b)²]

If we let C be the right angle, and let x be the length of the angle bisector from C, we find then that

x²=2a²b²/(a+b)².

Note that x cannot be an integer.

Let y and z be the lengths of the angle bisectors from A and B, respectively.

y²=bc[1 - a²/(b+c)²]

and

z²=ca[1 - b²/(c+a)²].

It suffices to show that at most one of y and z can be integral. The fact that c²=a²+b² should be used.