Prove that if you draw a right triangle and then draw a circle with its center on the hypotenuse's midpoint such that it intersects at least 1 vertex, it will in fact intersect all three.
Any right triangle can be placed on a Cartesian coordinate system such that the vertex between the two legs (opposite the hypotenuse) is at the origin.
Then the two other vertices can fall on (0,Y) and (X,0) respectively.
Clearly at the midpoint of the hypotenuse (which occurs at (X/2, Y/2) ) , the distance is exactly half way between (X,0) and (0,Y)...
And, if I draw the line segment from (X/2, Y/2) to the origin, then I have either of two iscocoles triangles, which shows that it is the same distance as to either other vertex.
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So, all vertices are equidistant (equal radius) from the midpoint of the hypotenuse.
Edited on November 14, 2003, 3:25 pm
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