Inverse Pythagorean theorem

Let A, B be the endpoints of the hypotenuse of a right triangle △ABC. Let D be the foot of a perpendicular dropped from C, the vertex of the right angle, to the hypotenuse. Then

${\displaystyle \frac{1}{C{D}^2}=\frac{1}{A{C}^2}+\frac{1}{B{C}^2}.}$

This theorem should not be confused with proposition 48 in book 1 of Euclid's Elements, the converse of the Pythagorean theorem, which states that if the square on one side of a triangle is equal to the sum of the squares on the other two sides then the other two sides contain a right angle.

Proof

The area of triangle △ABC can be expressed in terms of either AC and BC, or AB and CD:

${\displaystyle \begin{array}{cc}\frac{1}{2}AC\cdot BC& =\frac{1}{2}AB\cdot CD\\ (AC\cdot BC{)}^2& =(AB\cdot CD{)}^2\\ \frac{1}{C{D}^2}& =\frac{A{B}^2}{A{C}^2\cdot B{C}^2}\end{array}}$

given CD > 0, AC > 0 and BC > 0.

Using the Pythagorean theorem,

${\displaystyle \begin{array}{cc}\frac{1}{C{D}^2}& =\frac{B{C}^2+A{C}^2}{A{C}^2\cdot B{C}^2}\\ & =\frac{B{C}^2}{A{C}^2\cdot B{C}^2}+\frac{A{C}^2}{A{C}^2\cdot B{C}^2}\\ \therefore \frac{1}{C{D}^2}& =\frac{1}{A{C}^2}+\frac{1}{B{C}^2}\end{array}}$

as above.

Note in particular:

${\displaystyle \begin{array}{cc}\frac{1}{2}AC\cdot BC& =\frac{1}{2}AB\cdot CD\\ CD& =\frac{AC\cdot BC}{AB}\end{array}}$