Proving that the median to the hypotenuse of a right triangle is equal to half of the hypotenuse

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If I have a right triangle, of the form: $ABC$, with right angle at B:

Then, $BM = CM = AM$.

Then, as I show that:

$BM = AM$

The simplest demonstration is to create another triangle equal to this one, and adjust them so that it looks like a rectangle. The diagonals of a rectangle are the same, which shows the property.

But I would like to see the demonstration, without using the property of diagonals of a rectangle, in advance thanks.

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2 Answers

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Drop the altitude from $M$ to $AB$ which will be parallel to $BC$ and find two equal right triangles.

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Consider the circumcircle of ABC. Since ABC=90º, AC is the diameter.

Thus M is the circumcenter and MC=MB=MA.

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