Question

solve this if U dare ........ if the diagonals of a parallelogram are equal then show that it is a rectangle ¿¿...................?​

Answer 1

Answer:

Step-by-step explanation:

lets say, ABCD is a parallelogram



Given that the diagonals AC and BD of parallelogram ABCD are equal in length .

Consider triangles ABD and ACD.

AC = BD [Given]

AB = DC [opposite sides of a parallelogram]

AD = AD [Common side]

∴ ΔABD ≅ ΔDCA [SSS congruence criterion]

∠BAD = ∠CDA [CPCT]

∠BAD + ∠CDA = 180° [Adjacent angles of a parallelogram are supplementary.]

So, ∠BAD and ∠CDA are right angles as they are congruent and supplementary.

Therefore, parallelogram ABCD is a rectangle since a parallelogram with one right interior angle is a rectangle.

Answer 2

Given: ABCD is a parallelogram and AC = BD

To prove: ABCD is a rectangle

Proof:  In  Δ ACB and ΔDCB

AB = DC _____ Opposite sides of parallelogram are equal

BC = BC _____ Common side

AC = DB _____ Given

Therefore,

Δ ACB ≅ ΔDCB by S.S.S test

Angle ABC = Angle DCB ______ C.A.C.T

Now,

AB ║ DC _______ Opposite sides of parallelogram are parallel

Therefore,

Angle B + Angle C = 180 degree (Interior angles are supplementary)

Angle B + Angle B = 180

2 Angle B  = 180 degree

Angle B = 90 degree

Similarly, we can prove that, Angle A = 90 degree, Angle C = 90 degree and Angle D = 90 degree.

Therefore, ABCD is a rectangle.

(Refer to the attachment for the figure)