Proving the Equidistance of the Midpoint of the Hypotenuse in a Right Triangle

The focus of this article is to provide a detailed proof of the property that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices, using vector methods. This proof not only validates a specific geometric property but also leads to a converse and a fundamental theorem concerning inscribed angles.

Basis of Proof Using Vectors

To approach this problem, we first set up the vectors for the vertices and midpoint:

(overrightarrow{CA} mathbf{v}) (overrightarrow{CB} mathbf{w}) (overrightarrow{CM} mathbf{m}) (overrightarrow{MA} mathbf{h}) Since (M) is the midpoint of (overline{AB}), (overrightarrow{MB} -mathbf{h})

With these assignments, we can write:

(mathbf{m} mathbf{v} mathbf{h})

(mathbf{m} mathbf{w} - mathbf{h})

From these, we derive:

(mathbf{m} cdot mathbf{h} mathbf{v})

(mathbf{m} cdot (-mathbf{h}) mathbf{w})

Bringing these into a specific expression:

(|mathbf{m}|^2 - |mathbf{h}|^2 mathbf{v} cdot mathbf{w})

Thus,

(|mathbf{m}| sqrt{|mathbf{h}|^2 mathbf{v} cdot mathbf{w}})

Specializing to a Right Triangle

When triangle (ACB) is a right triangle with the right angle at (C), (mathbf{v} cdot mathbf{w} 0) because the vectors (mathbf{v}) and (mathbf{w}) are perpendicular. Therefore, we have:

(|mathbf{m}| sqrt{|mathbf{h}|^2 0} |mathbf{h}|)

This proves that the midpoint (M) of the hypotenuse (overline{AB}) is equidistant from the three vertices.

Converse and Extended Theorems

Not only does this proof establish the property for a right triangle, but it also provides a converse and links to an important theorem about inscribed angles:

Converse Proposition

If the median (overline{MC}) in triangle (ABC) is the same length as (overline{MA}) and (overline{MB}), then triangle (ABC) is a right triangle with the right angle at (C).

Proof: If (|mathbf{m}| |mathbf{h}|), the expression forces (mathbf{v} cdot mathbf{w} 0), which is the condition for a right angle.

Corollary: Inscribed Angle Theorem

Any angle inscribed in a semicircle is a right angle.

Proof: An angle inscribed in a semicircle is such that the midpoint (M) of (overline{AB}) is the center of the semicircle, and points (A), (B), and (C) lie on the circumference. Since (overline{MA}), (overline{MB}), and (overline{MC}) are radii of the semicircle, they are all of equal length. The proposition then states that (triangle ABC) is a right triangle with the right angle at (C).