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This question is related to one of my previous questions.

The answer to that question included a theorem: "The median on the hypotenuse of a right triangle equals one-half the hypotenuse".

When I wrote the answer out and showed it a friend of mine, he basically asked me how I knew that the theorem was true, and if the theorem had a name.

So, my question:

-Does this theorem have a name?
-If not, what would be the best way to describe it during a math test? Or is it better to write out the full prove every time?

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    One could call it the converse of the [Theorem of Thales.](http://en.wikipedia.org/wiki/Thales%27_theorem)2012-11-19

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Here is a proof without words:

enter image description here

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If you are interested in seeing proofs: see these proofs. You'll actually find two proofs (and illustrations) of the following theorems:

  1. If a triangle is a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse (from which it follows that the median drawn to the hypotenuse divides the triangle in two isosceles triangles); and

  2. If in a triangle a median has the measure half the length of the side it is drawn, then the triangle is a right triangle.


The proofs are not at all elaborate, and utilize properties you already know, so they are easy enough to reconstruct, if necessity dictates that you do so. Understanding "why" these theorems hold is the important point; correctly "naming" them is a less important.

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    @Babak: you're worth ${\small\infty}^{\Large\infty}$2013-06-09
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You could call it a special case of Apollonius's theorem, or of the parallelogram law.

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    This is exactly how I remember it - as a special case of Appolonius theorem :-)2012-11-19
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Let us consider the right triangle ABC with the right angle  and let AD be the median drawn from the vertex A to the hypotenuse  BC. We need to prove that the length of the median AD is half the  length of the hypotenuse BC. 

Draw the straight line DE passing through the midpoint D parallel to the  leg AC till the intersection with the other leg AB at the point E. The angle BAC is the right angle by the condition. The angles BED and BAC are congruent as they are corresponding angles at the parallel lines AC and ED and the transverse. AB Therefore, the angle BED is the right angle. 

Now, since the straight line DE passes through the midpoint D and is parallel to AC, it cuts the side AB in two congruent segments of equal length: AE = EB  So, the triangles AED and BED are right triangles that have congruent  AE and EB and the common DE. Hence, these triangles are congruent in accordance to the postulate (SAS)  It implies that the segments AD and DB are congruent as corresponding sides of these triangles. Since DB has the length half the length of the hypotenuse BC, we have proved that the median AD has the length half the length of the hypotenuse. 

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In a circle, a triangle which has for side a diameter is rectangle. This means the center of the circle is the mean point of the diameter, and the median is a radius of the circle.

You can prove this as a corollary to the fact that inscribed angles are half of the center ones. So I guess that is a corollary to this corollary.