Complete the following proof.

prove: the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.

(fill in the blanks of the equation in the second picture with the correct number/letter/sign based off the first picture.)

Prove CM = BM = AM Again, I'll do it, you can fill in the blanks. You almost have to make a print out for these questions.

CM
C = (0,b)
M = (a/2,b/2)
CM^2 = (0 - a/2 )^2 + (b - b/2)^2
CM^2 = a^2 / 4 + (b/2)^2
CM^2 = a^2 / 4 + b^2 / 4 

AM
A = (0,0)
M = (a/2,b/2)
AM^2 = (a/2 - 0)^2 + (b/2 - 0)^2
AM^2 = a^2 / 4 + b^2 / 4

Well what a piece of luck. We don't need to find the relationship between a and b. I check later to see if I need to do that.

BM
B = (a,0)
M = (a/2, b/2)
BM^2 = (a - a/2)^2 + ( 0 - b/2)^2
BM^2 = (a/2)^2 + (- b/2)^2
BM^2 = a^2/4 + b^2/4 and again we are done.