Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Advertisement Remove all ads
Solution
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.
Concept: Another Condition for a Quadrilateral to Be a Parallelogram
Is there an error in this question or solution?