Q3 of 33 Page 108

Let us prove that, a parallelogram whose diagonals intersect at right angles is a rhombus.


Consider the parallelogram ABCD with diagonals AC and BD as shown and they intersect at right angles at O


A parallelogram is a rhombus if its adjacent sides are equal


Consider ΔAOB and ΔAOD


∠AOB = ∠AOD … both 90° because given that diagonals intersect at right angles


OD = OB … diagonals of a parallelogram bisect each other


AO is the common side


Therefore, ΔAOB ≅ ΔAOD … SAS test for congruency


⇒ AB = AD … corresponding sides of congruent triangles


Thus, adjacent sides are equal


Thus, we can conclude that parallelogram ABCD is a rhombus


Therefore, a parallelogram whose diagonals intersect at right angles is a rhombus.


More from this chapter

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1

Let us prove that, if the lengths of two diagonals of a parallelogram are equal, then the parallelogram will be a rectangle.

 2

Let us prove that, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.

 4

The two diagonals of a parallelogram ABCD intersect each other at the point O. A straight line passing through O intersects the sides AB and DC at the points P and Q respectively. Let us prove that OP = OQ.

 5

Let us prove that in an isosceles trapezium the two angles adjacent to any parallel sides are equal.