Chapter P.2, Problem 65E

Solution

To prove that the midpoint of the hypotenuse of any right triangle is equidistant from the three vertices.

Given:

A right triangle.

Calculation:

Let ACB be a right triangle with right angle at C . Let the length of legs be a and b , and the length of hypotenuse be c .

Let the vertex C be at the origin $\left(0,0\right)$ , then the coordinates of the vertices B and A be $\left(a,0\right)\text{ and }\left(0,b\right)$ respectively.

The midpoint of the hypotenuse is given by

  $\left(\frac{a+0}{2},\frac{0+b}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)$

And in the right-angle triangle ACB , using the Pythagorean theorem, it follows

  $a^2+b^2=c^2$

The midpoint of the hypotenuse is $\frac{c}{2}$ units from the vertices A and B .

Find the distance of the midpoint $\left(\frac{a}{2},\frac{b}{2}\right)$ from the vertex C $\left(0,0\right)$ using the distance formula as shown below

  $\begin{array}{c}\sqrt{{\left(\frac{a}{2}-0\right)}^2+{\left(\frac{b}{2}-0\right)}^2}=\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}\\ =\sqrt{\frac{a^2+b^2}{4}}\\ =\sqrt{\frac{c^2}{4}}\left(\text{since}{a}^2+b^2=c^2\right)\\ =\frac{c}{2}\end{array}$

It implies that the midpoint of the hypotenuse of any right triangle is equidistant from all three vertices.