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Question

If the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a:

A
square
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B
rhombus
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C
rectangle
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D
kite
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Solution

The correct option is A square

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC=BD, OA=OC, OB=OD, and AOB=BOC=COD=AOD=90. To prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB=BC=CD=AD, and one of its interior angles is 90.

In ΔAOB and ΔCOD,
AO=CO (Diagonals bisect each other)
OB=OD (Diagonals bisect each other)
AOB=COD Vertically opposite angles)
ΔAOBΔCOD (SAS congruence rule)
AB=CD (By CPCT) …(1)
And OAB=OCD (By CPCT)
However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
AB||CD .…(2)
From equations (1) and (2). We obtain that, ABCD is a parallelogram.

In ΔAOD and ΔCOD
ΔAODΔCOD (SAS congruence rule)
AD = DC…..(3)
However, AD=BC and AB=CD (Opposite sides of parallelogram ABCD)
AB =BC=CD=DA
Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ΔADC and ΔBCD,
AD=BC (Already proved)
AC=BD (Given)
DC=CD (Common)
ΔADCΔBCD (SSS Congruence rule)
ADC=BCD (By CPCT)
However, ADC+BCD=180 (Co-interior angles)
ADC+ADC=180
2ADC=180
ADC=90
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram, AB=BC=CD=AC and one of its interior angles is 90.
Therefore, ABCD is a square.

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