Prove that the diagonals of a rectangle bisect each other and are equal using co-ordinate geometry.
Put the lower left corner at the origin A(0, 0) and
let the other 3 corners clockwise be B(0,b), C(a, b), D(a, 0). Label as indicated.
Equality:
AC² : (a–0)² + (b–0)² = a²+b²
BD²: (0–a)² + (b–0² = 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 !!
