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

Asked by ISHAN BHANDARI 16th December 2011, 12:00 AM
Answered by Expert
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. 


AC : (a0) + (b0) = a+b
BD: (0a) + (b0 = a+b

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 !! 
Answered by Expert 19th December 2011, 12:10 AM
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