1) Show that the circumradius of a right triangle is equal to half the hypotenuse.
Let ABC be a right triangle and let B be the right angle
Let B = (0,0), A = (0, a) and C = (c ,0)
So......the midpoint of the hypotenuse is given by (c/2, a/2)
And the distance ffrom this point to A is √ [ c^2 /4 + a^2/4] = √ [ a^2 + c^2] / 2
And the distance from this point to C is √ [ c^2/4 + + a^2/4 ] = √ [ a^2 + c^2] / 2
And the distance from this point to B is √ [ a^2 + c^2] / 2
So.....a circle centered at the midpoint of the hypotenuse will pass through all three of the vertices since they are all equally distant from this midpoint