Question

Prove that in a rectangle the diagonals bisect each other and equal

Answer 1

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 !! 

Answer 2

Your Question :

Prove that in a rectangle the diagonals bisect each other and equal

Given Information :

In the adjoining figure, the diagonals AC and BD of the rectangle ABCD intersect at the point O.

To Prove :

• OA = OC and OB = OD
• OA = OB = OC = OD
• Diagonal AC = Diagonal BD

Proof :

In ∆OAB and ∆OCD,

➻ ∠OAB = ∠OCD ㅤㅤ (∵ AB || DC and alternate angles are equal)

➻ AB = DC ㅤㅤ (∵ opposite sides are equal)

And ∠OBA = ∠ODC ㅤㅤ (∵ AB || DC and alternate angles are equal)

∴ ∆OAB ≅ ∆OCD ㅤㅤ (A-S-A conditions of congruence)

So, the corresponding sides of ∆OAB and ∆OCD are equal.

∴ OA = OC and OB = OD.

But OA = 1/2AC, OB = 1/2BD and AC = BD.

Hence, OA = OB = OC = OD ㅤㅤ [Proved]

Now, In the figure above,

∆ABC ≅ ∆BAD, because AB = BA, ∠ABC = ∠BAD = 90° and BC = AD.

So, diagonal AC = diagonal BD.

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