Proving the Midpoints of the Hypotenuse in a Right Triangle are Equidistant from All Vertices

In the realm of Euclidean geometry, one fascinating observation is that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices. This property can be proven through both geometric reasoning and the application of the distance formula. Here, we will walk through the steps to prove this statement, ensuring that the explanation is clear and accessible.

Step 1: Setting Up the Triangle

Consider a right triangle (triangle ABC) on the XY plane, with vertices at the following coordinates:

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

The hypotenuse BC connects points B and C. Let's proceed to find the midpoint of the hypotenuse, M.

Step 2: Finding the Midpoint of the Hypotenuse

The midpoint of a line segment is the average of the coordinates of its endpoints. For the hypotenuse BC, the midpoint M can be calculated as:

[text{M} left(frac{x_B x_C}{2}, frac{y_B y_C}{2}right) left(frac{a 0}{2}, frac{0 b}{2}right) left(frac{a}{2}, frac{b}{2}right)]

Step 3: Calculating Distances from Midpoint to Vertices

To prove that M is equidistant from all vertices, we will calculate the distances from M to A, B, and C using the distance formula:

[text{Distance} sqrt{(x_2 - x_1)^2 (y_2 - y_1)^2}] Distance from M to A: [text{d}_text{M}A sqrt{left(frac{a}{2} - 0right)^2 left(frac{b}{2} - bright)^2} sqrt{left(frac{a}{2}right)^2 left(-frac{b}{2}right)^2} sqrt{frac{a^2}{4} frac{b^2}{4}} frac{1}{2}sqrt{a^2 b^2}] Distance from M to B: [text{d}_text{M}B sqrt{left(frac{a}{2} - aright)^2 left(frac{b}{2} - 0right)^2} sqrt{left(-frac{a}{2}right)^2 left(frac{b}{2}right)^2} sqrt{frac{a^2}{4} frac{b^2}{4}} frac{1}{2}sqrt{a^2 b^2}] Distance from M to C: [text{d}_text{M}C sqrt{left(frac{a}{2} - 0right)^2 left(frac{b}{2} - bright)^2} sqrt{left(frac{a}{2}right)^2 left(-frac{b}{2}right)^2} sqrt{frac{a^2}{4} frac{b^2}{4}} frac{1}{2}sqrt{a^2 b^2}]

Step 4: Conclusion

We have shown that:

[text{d}_text{M}A text{d}_text{M}B text{d}_text{M}C frac{1}{2}sqrt{a^2 b^2}]

Hence, the midpoint M of the hypotenuse BC is equidistant from all three vertices A, B, and C. This confirms the statement and provides a clear geometric proof of the property.