Question

prove that diagonals of rectangle are equal and bisect each other​

Answer 1

Question:

• To prove that diagonals of a rectangle are equal and bisect each other.

Answer:

• Let us take a rectangle ABCD and there diagonals AC and BC intersect at the point O.

From ∆ABC and ∆BAD ,we have

AB = BA

∠ABC = ∠BAD

BC = AD

∆ABC ≅ ∆BAD

AC = BD

Thus,the diagonals of the rectangle are equal.

From ∆OAB & ∆OCD ,we have

∠OAB = ∠OCD

∠OBA = ∠ODC

AB = CD

∆OAB ≅ ∆OCD

OA = OC & OB = OD

Hence, this shows the diagonals of rectangle bisect each other.

Therefore, it is proved that the diagonals of a rectangle are equal and also bisect each other.

Answer 2

Let ABCD be the required rectangle. Let the diagonals be AC and BD. Let the intersect each other at O.

In △ABD and △BAC,

1. AD = BC (opposite sides of a rect. are same),
2. ∠BAD = ∠ABC = 90° (angles of a rect. is equal to 90°), and
3. AB = BA (common).

∴△ABD ≅ △BAC by S–A–S congruence criterion.

So,

∠ADB = ∠BCA (by c.p.c.t.)

In △AOB and ∠BOC,

1. ∠AOB = ∠BOC (vertically opposite angles),
2. AD = BC (opposite sides are same), and
3. ∠ADB = ∠BCA (already proved)

∴△AOB ≅ △BOC by A–S–A congruence criterion.

∴ We can conclude that,

• AC = BD, (by c.p.c.t. of Case 1)
• AO = CO and DO = BO, (By c.p.c.t. of Case 2.)

So, the diagonals of a rectangle bisect each other and are congruent.