+2448 
It's a bit messy but im sure you get what im goin at here. This is what i've got so far:
1. Given
2. AB= \(3^2+1^2=c^2\)
9+1=c^2
sqrt(10)=sqrt(c^2)
c=sqrt(10)
BC=\(3^2+1^2=c^2\)
9+1=c^2
sqrt(10)=sqrt(c^2)
c=sqrt(10)
CD=\(3^2+1^2=c^2\)
9+1=c^2
sqrt(10)=sqrt(c^2)
c=sqrt(10)
DA=\(3^2+1^2=c^2\)
9+1=c^2
sqrt(10)=sqrt(c^2)
c=sqrt(10)
3. \(AB\cong BC\cong CD \cong DA\)
4. I'm stuck here >.<
I know that they are congruent but how do i prove that and i know that it is obviously a square
+128475 Here's the way I might do it, RP
1. Find the side lengths
2. Show that the diagonals have the same length
3. Show that the length of any side =
diagonal length / √2
3. This will mean that we have a square because the sides are congruent, the diagonals are congruent and because of the proof of (3), they will meet at right angles
+128475 Example .....to find a side length.....use the distance formula
Are you familiar with this???
Distance from A to B =
sqrt [ ( 1 - 2)^2 + (3 - 0)^2 ] = sqrt [ (-1)^2 + 3^2] = sqrt [ 1 + 9 ] = sqrt (10)
Then find the distances between B,C C,D and D,A
They should all be = sqrt (10)
Let me know if you have trouble
