Proving the Midpoint of Hypotenuse as Circumcenter and Determining the Circumradius in a Right Triangle

A right triangle is a fundamental shape in geometry, characterized by one of its angles measuring 90°. The circumcenter of a triangle is the point where the perpendicular bisectors of its sides intersect. In the case of a right triangle, the midpoint of the hypotenuse serves as the circumcenter, and the distance from the right vertex to this midpoint is the circumradius. Let's explore how to prove this.

Definitions

Right Triangle: A triangle with one angle measuring 90°. Circumcenter: The point where the perpendicular bisectors of the sides of a triangle intersect. It is also the center of the circumcircle, the circle that passes through all the vertices of the triangle. Circumradius: The radius of the circumcircle of the triangle, which is the distance from the circumcenter to any vertex.

Setup

Let triangle ABC be a right triangle with C as the right angle, and AB as the hypotenuse. Let M be the midpoint of segment AB. We will prove that M is the circumcenter and that the line segment joining the right vertex to the midpoint of the hypotenuse is the circumradius.

Step 1: Show that M is the Circumcenter

Assume the coordinates of the vertices are as follows:

A(0, 0) B(a, 0) C(0, b)

The coordinates of M, the midpoint of AB, are:

[Mleft(frac{a}{2}, 0right)]

Let's calculate the distances from M to each vertex:

Distance MA: [MA sqrt{left(frac{a}{2} - 0right)^2 (0 - 0)^2} frac{a}{2}] Distance MB: [MB sqrt{left(frac{a}{2} - aright)^2 (0 - 0)^2} frac{a}{2}] Distance MC: [MC sqrt{left(frac{a}{2} - 0right)^2 (0 - b)^2} sqrt{left(frac{a}{2}right)^2 b^2}]

Using the Pythagorean Theorem

Since C is the right angle, we have:

[AC^2 BC^2 AB^2 implies b^2 a^2 AB^2]

This equation shows that:

[sqrt{left(frac{a}{2}right)^2 b^2} frac{sqrt{a^2 b^2}}{2} frac{AB}{2}]

Therefore, MA MB, and M lies on the perpendicular bisector of AB. Since the perpendicular bisector of AB is vertical and passes through M, M is equidistant from A, B, and C, making it the circumcenter.

Step 2: Circumradius

The circumradius R is the distance from M to C:

[R MC sqrt{left(frac{a}{2}right)^2 b^2}]

Simplifying using the Pythagorean theorem, we get:

[R frac{1}{2}sqrt{a^2 b^2} frac{1}{2}AB]

Where AB is the length of the hypotenuse.

Conclusion

We have shown that:

The midpoint M of the hypotenuse AB is the circumcenter of triangle ABC. The distance from the right vertex C to the midpoint M is the circumradius.

Hence the statement is proven.