Chapter 13, Problem 28RP

To determine

To Prove: The midpoint of the hypotenuse of any right triangle is equidistant from the vertices of the triangle.

Explanation of Solution

Given:

The given points are $\left(0,0\right),\left(2a,0\right),\left(0,2b\right)$ .

Calculation:

The midpoint of the vertices is obtained as,

  $\begin{array}{c}M=\left(\frac{2a+0}{2},\frac{2b+0}{2}\right)\\ =\left(a,b\right)\end{array}$

Then, the distance from mid-point to each of the vertex is,

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

Hence, it is proved that midpoint of the hypotenuse of any right triangle is equidistant from the vertices of the triangle.