Question 426510
The midpoint of the hypotenuse BC is ({{{a/2}}}, {{{b/2}}}). We can find the distance from each point to the midpoint and show that they are all equal.


A better way to prove this is to circumscribe a circle around ABC. Since BAC is a right angle, the hypotenuse BC is a diameter of the circle. We can draw the midpoint D and show that DA, DB, DC are all radii of the circle, hence they are equal.