Prove the median of right triangle is half the length of the hypotenuse

Question

How can I prove that in a right triangle the median which tends to the hypotenuse has length half of that the hypotenuse? I want to show using vector methods or analytic geometry

Answer 1

This follows from a result about triangles inscribed in circles and angles subtended by chords.

Any chord of a circle subtends a fixed angle to any point on the circle on the acute side, and another fixed angle to any point on the obtuse side. These two angles sum to pi radians/180 degrees.

By a symmetry argument, the chord subtends an right angle on both sides if and only if it is a diameter of the circle.

Any triangle can have a circle circumscribed through its vertices. Where the triangle is right angled, the hypotenuse naturally subtends a right angle at the third point, and thus is a diameter of the circle. The median of the triangle is a radius of the circle and thus half the hypotenuse.

Answer 2

You can place the triangle with vertices $(0,0), (0,a), (b,0)$, find the point where the median hits the hypotenuse, compute the length of the median, and compare to the length of the hypotenuse.

Answer 3

The converse can be proved to same logic at first.

Let $A$ be half hypotenuse and $B$ be the median vectors

Assume $ |A| = |B|= u $

Taking dot scalar product of sides

$$ ( A + B).(A-B) = |A|^2 - |B|^2= 0 \tag{1} $$

So the sides are perpendicular, must belong to a right triangle.

Next, for the direct proposition by the above relation (1) median length (radius) is half of the hypotenuse.