Use vectors and the dot product to prove that the midpoint of the hypotenuse of a right triangle is equidistant to all three vertices.
I know the dot product is A⋅B = |A||B|cosΘ ...... or .... A1B1 + A2B2 + A3B3 ... + AnBn
I know the magnitude of a vector is its length, and is given by√(x^2 + y^2) I have omitted the z^2 since I want to solve this in R^2.
The Attempt at a Solution
I figure doing this in R^2 is easier, so I drew a right triangle with points P(0,0), R(0, y), S(x, 0)
I know the midpoint would be given by T = (x/2, y/2)
I constructed vectors using these points:
PS = <x, 0>
PR = <0, y>
TS = <x/2, -y/2>
RT = <x/2, -y/2>
PT = <x/2, y/2>
From this I can show that the magnitude of TS, RT, and PT are all the same, thus their distances are the same. I can also use this to show that TS = RT. I can then take the dot product of TS and RT and work it down to show the angle between them is 0 (so cosΘ = 1) and thus they are the same vector.
What I cannot do for the life of me is use the dot product to prove that PT has the same magnitude as the other two vectors, TS and RT. I just can't figure it out. I've tried finding ways to express PT as other vectors but everything I've tried has failed. I am tearing my hair out!
Here is a diagram of how I am trying to layout my triangle:
I would greatly appreciate any insight into this very troubling problem. :(