Proving the Midpoint of the Hypotenuse in a Right-Angled Triangle is Equidistant from the Vertices

In this article, we will explore the vector method to prove a fundamental property of right-angled triangles. Specifically, we will show that the midpoint of the hypotenuse is equidistant from the vertices of the triangle. This not only establishes a deep geometric relationship but also leads to a well-known theorem about inscribed angles in semicircles.

Introduction to the Problem

Consider a right-angled triangle ( triangle ABC ), where ( angle ACB ) is the right angle. It is well known that the hypotenuse of this triangle is the diagonal of a rectangle when divided from the right angle. This rectangle's diagonal not only bisects the right angle but also connects the vertices of the rectangle. The midpoint of this diagonal—equivalently, the hypotenuse of ( triangle ABC )—is equidistant from all four vertices of the rectangle. Therefore, it is also the circumcenter of ( triangle ABC ).

Using Vector Methods

Let's use vector methods to prove this property. We will assign vectors to the vertices of the triangle and use vector operations to derive the necessary relationships.

The Setup

Let ( overrightarrow{CA} mathbf{v} ) and ( overrightarrow{CB} mathbf{w} ). The midpoint ( M ) of ( overline{AB} ) is denoted by ( overrightarrow{CM} mathbf{m} ) and ( overrightarrow{MA} mathbf{h} ). Since ( M ) is the midpoint of ( overline{AB} ), we have ( overrightarrow{MB} -mathbf{h} ).

From the vector assignments:

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

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

Combining these equations, we can express ( mathbf{h} ) in terms of ( mathbf{m} ), ( mathbf{v} ), and ( mathbf{w} ).

Deriving the Midpoint Distance

To show that the midpoint ( M ) is equidistant from ( A ), ( B ), and ( C ), we need to prove ( |mathbf{m}| |mathbf{h}| ). Let's start with the vector equation:

[ mathbf{m} cdot mathbf{h} mathbf{v} cdot mathbf{w} ]

Expanding this, we get:

[ |mathbf{m}|^2 - |mathbf{h}|^2 mathbf{v} cdot mathbf{w} ]

Since ( triangle ABC ) is a right-angled triangle at ( C ), ( mathbf{v} cdot mathbf{w} 0 ). Therefore:

[ |mathbf{m}|^2 |mathbf{h}|^2 ]

Thus, ( |mathbf{m}| |mathbf{h}| ), proving that the midpoint of the hypotenuse is equidistant from the vertices of the triangle.

Generalization and Converse

The vector method we used is not limited to right-angled triangles. For any triangle, if ( M ) is the midpoint of ( overline{AB} ), then:

[ |mathbf{m}| sqrt{|mathbf{h}|^2 mathbf{v} cdot mathbf{w}} ]

If the triangle is right-angled at ( C ), then ( mathbf{v} cdot mathbf{w} 0 ), and the above expression simplifies to:

[ |mathbf{m}| |mathbf{h}| ]

This generalizes to a converse statement:

Converse: Right Triangle Condition

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 of the Converse

Given ( |mathbf{m}| |mathbf{h}| ), we have:

[ mathbf{v} cdot mathbf{w} 0 ]

This implies that ( mathbf{v} ) and ( mathbf{w} ) are orthogonal, which is the condition for ( angle ACB ) to be a right angle.

Corollary: Inscribed Angle in a Semicircle

As a corollary, we can derive the well-known theorem that an angle inscribed in a semicircle is a right angle. Consider a semicircle with diameter ( overline{AB} ) and center ( M ). If ( M ) is the midpoint of the hypotenuse of a right-angled triangle ( triangle ABC ), then ( overline{MA} ), ( overline{MB} ), and ( overline{MC} ) are all radii of the semicircle, and therefore equal in length.

Proof of the Corollary

Since ( M ) is the center of the semicircle and ( overline{MA} ), ( overline{MB} ), and ( overline{MC} ) are all radii, we have ( |mathbf{m}| |mathbf{h}| |mathbf{v}| ). From the earlier proof, ( |mathbf{m}| |mathbf{h}| ) implies ( mathbf{v} cdot mathbf{w} 0 ), which means ( angle ACB ) is a right angle.

This completes the proof and deepens our understanding of the geometric properties of right-angled triangles and their relationships with vectors and semicircle theorems.