Given: A(2a, 0), B(0, 2b) and C(0, 0). Prove that the midpoint of the hypotenuse of right ΔABC is equidistant from vertices A, B, and C.

Question

Given: A(2a, 0), B(0, 2b) and C(0, 0). Prove that the midpoint of the hypotenuse of

right ΔABC is equidistant from vertices A, B, and C.

Answer 1

The midpoint of the hypotenuse is (a,b)...and the easiest way to prove this is:

if the co-ordinates of a line segment are (a,b) and (c,d), the co-ordinates of the midpoint are [(a+c)/2,(b+d)/2]

Also the so-called side-splitter theorem will give you the same result.

Then there are 4 congruent right triangles...draw a figure and you will immediately see this.