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prove that if the diagonals of a parallelogram are equal,then its a rectangle

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

∴ ΔABC ≅ ΔDCB (By SSS Congruence rule)

⇒ ∠ABC = ∠DCB

It is known that the sum of the measures of angles on the same side of transversal is 180º.

∠ABC + ∠DCB = 180º (AB || CD)

⇒ ∠ABC + ∠ABC = 180º

⇒ 2∠ABC = 180º

⇒ ∠ABC = 90º

Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.

 

  

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View Full Answer

LET THE QUADRILATERAL BE ABCD .

AC = BD .

AB = AB

AD = BC

SO , TRIANGLE ABD IS CONGRUENT TO TRIANGLE BAC

SO , ANGLE A IS EQUAL TO ANGLE B ( BY CPCT )

BUT ANGLE A + ANGLE B = 180

SO , ANGLE A IS 90

ANGLE B = 90

AS ANGLE A + ANGLE  = 180

ANGLE D = 180

ANGLE A = ANGLE C

ANGLE C = 90

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