Q. Prove that the diagonals of a rectangle bisect each other and one equal?
let the other 3 corners clockwise be B(0,b), C(a, b), D(a, 0). Label as indicated.
AC² : (a0)² + (b0)² = a²+b²
BD²: (0a)² + (b0² = a²+b²
Thus AC = BD.
Bisection. Compute the midpoints of each and show itis the same point for each diagonal.
Midpoint of AC: ((0+a)/2, (0+b)/2) = (a/2, b/2)
Midpoint of BD: ((0+a)/2, (b+0)/2) = (a/2, b/2)
Since the midpoints are common, that is where they cross. Since where they cross are the midponts, the diagonals bisect each other. HENCE PROVED !!