Question

21.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.​

Answer 1

Given

Parallelogram ABCD

AC = BD

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To Prove

The parallelogram ABCD is a rectangle.

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Solution

We know that the interior angles of a rectangle is 90°

So we need to find one interior angle in this parallelogram which is 90°.

So we'll divide this parallelogram into two triangles. Namely ΔABC and ΔDCB.

The opposite sides of a parallelogram is always equal.

∴ AB = DC

BC = BC since they are common.

And we are given that AC = DB

ΔABC ≅ ΔDCB by the SSS Congruence

∠ABC = ∠DCB

AB is parallel to DC and BC is the transversal.

∠B + ∠C = 180°

B + B = 180°

2B = 180°

B = 180 ÷ 2 = 90°

∴ We found that one of the interior angle of the parallelogram is 90°

∴ Hence it's proved that the above is rectangle ABCD.

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Answer 2

Step-by-step explanation:

Gven: In parallelogram ABCD, AC=BD

To prove : Parallelogram ABCD is rectangle.

Proof : in △ACB and △BDA

AC=BD ∣ Given

AB=BA ∣ Common

BC=AD ∣ Opposite sides of the parallelogram ABCD

△ACB ≅△BDA∣SSS Rule

∴∠ABC=∠BAD...(1) CPCT

Again AD ∥ ∣ Opposite sides of parallelogram ABCD

AD ∥BC and the traversal AB intersects them.

∴∠BAD+∠ABC=180∘

...(2) _ Sum of consecutive interior angles on the same side of the transversal is

180∘

From (1) and (2) ,

∠BAD=∠ABC=90∘

∴∠A=90∘

and ∠C=90∘

Parallelogram ABCD is a rectangle.

and ∠C=90∘

Parallelogram ABCD is a rectangle.