Question

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
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Answer 1

$\huge \underline \mathbb {SOLUTION:-}$

Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

• Figure provided in the above attachment.

Therefore:

• In ∆AOB and ∆AOD

We have:

• AO = AO [Common]
• OB = OD [O is the mid-point of BD]
• ∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

Therefore:

AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC ...(2)

BC = CD …..(3)

CD = DA ……(4)

Therefore:

From (1), (2), (3) and (4), we have

• AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Alternatively: ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

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Answer 2

Step-by-step explanation:

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.

In △AOB and △AOD

DO=OB ∣ O is the midpoint

AO=AO ∣ Common side

∠AOB=∠AOD ∣ Right angle

So, △AOB≅△AOD

So, AB=AD

Similarly, AB=BC=CD=AD can be proved which means that ABCD is a rhombus.