Question

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.)

Answer 1

`M=\left(\frac{0+\boxed{a}}{2},\frac{b+\boxed{0}}{2}\right)=\left(\frac{\boxed{a}}{2},\frac{\boxed{b}}{2}\right)\\\\\\`


`MB=\sqrt{\left((a)/(2)-\boxed{a}\right)^2+\left(\frac{\boxed{b}}{2}-\boxed{0}\right)^2}=\\\\\\= \sqrt{\left(\frac{a}{\boxed{2}}-\frac{\boxed{2}\boxed{a}}{2}\right)^2+\left(\frac{b}{\boxed{2}}\right)^2}=\\\\\\= \sqrt{\left(\frac{-\boxed{a}}{2}\right)^2+\left(\frac{b}{\boxed{2}}\right)^2}=\sqrt{\frac{a^2}{\boxed{4}}+\frac{b^2}{\boxed{4}}}`


`MC=\sqrt{\left((a)/(2)-\boxed{0}\right)^2+\left((b)/(2)\boxed{-}\boxed{b}\right)^2}=\\\\\\= \sqrt{\left(\frac{a}{\boxed{2}}\right)^2+\left(\frac{\boxed{b}}{2}-\frac{\boxed{2}\boxed{b}}{2}\right)^2}=\\\\\\= \sqrt{\left(\frac{a}{\boxed{2}}\right)^2+\left(-\frac{\boxed{b}}{2}\right)^2}=\sqrt{\frac{a^2}{\boxed{4}}+\frac{b^2}{\boxed{4}}}`


`MA=\sqrt{\left((a)/(2)-\boxed{0}\right)^2+\left(\frac{\boxed{b}}{2}-\boxed{0}\right)^2}=\\\\\\= \sqrt{\left(\frac{\boxed{a}}{2}\right)^2+\left(\frac{b}{\boxed{2}}\right)^2}=\sqrt{\frac{a^2}{\boxed{4}}+\frac{\boxed{b}^2}{\boxed{4}}}`

Answer 2

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.