Question

Show that the diagonals of a square are equal and bisect each other at right angles , then it is a rhombus​

Answer 1

Step-by-step explanation:

A triangle has three sides. Any one of the lines that form a triangle is called its side. This is the side at the bottom of the triangle. The base of a triangle is the side on which it “rests.”

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Answer 2

Step-by-step explanation:

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e.,

OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. To prove

ABCD a rhombus, we have to prove ABCD is a parallelogram and all the sides of ABCD are

equal.

In ΔAOD and ΔCOD,

OA = OC                      (Diagonals bisect each other)

∠AOD = ∠COD          (Given)

OD = OD                    (Common)

So, ΔAOD ≅ ΔCOD (By SAS congruence rule)

Hence, AD = CD …………..1

Similarly, it can be proved that

AD = AB and CD = BC ………..2

From equation 1 and 2, we get

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, it can be said that ABCD is a

parallelogram. Since all sides of a parallelogram ABCD are equal, it can be said that ABCD is a

rhombus.