Question
Fri December 16, 2011 By: Ishan Bhandari

Q. Prove that the diagonals of a rectangle bisect each other and one equal?

Expert Reply
Mon December 19, 2011
We prove that the diagonals of a rectangle bisect each other and are equal using co-ordinate geometry as:
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 !! 
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