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

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

$MB=\sqrt{\left(\dfrac{a}{2}-\boxed{a}\right)^2+\left(\dfrac{\boxed{b}}{2}-\boxed{0}\right)^2}=\\\\\\= \sqrt{\left(\dfrac{a}{\boxed{2}}-\dfrac{\boxed{2}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\\\\\\= \sqrt{\left(\dfrac{-\boxed{a}}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{a^2}{\boxed{4}}+\dfrac{b^2}{\boxed{4}}}$

$MC=\sqrt{\left(\dfrac{a}{2}-\boxed{0}\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{b}\right)^2}=\\\\\\= \sqrt{\left(\dfrac{a}{\boxed{2}}\right)^2+\left(\dfrac{\boxed{b}}{2}-\dfrac{\boxed{2}\boxed{b}}{2}\right)^2}=\\\\\\= \sqrt{\left(\dfrac{a}{\boxed{2}}\right)^2+\left(-\dfrac{\boxed{b}}{2}\right)^2}=\sqrt{\dfrac{a^2}{\boxed{4}}+\dfrac{b^2}{\boxed{4}}}$

$MA=\sqrt{\left(\dfrac{a}{2}-\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}-\boxed{0}\right)^2}=\\\\\\= \sqrt{\left(\dfrac{\boxed{a}}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

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.