Question

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

• A. $square$
• B. $rhombus$
• C. $rectangle$
• D. $kite$

Answer 1

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 $\angle AOB=\angle BOC=\angle COD=\angle AOD={90}^{\circ }$. 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 ${90}^{\circ }$.

In $\Delta AOBand\Delta COD,$
AO=CO (Diagonals bisect each other)
OB=OD (Diagonals bisect each other)
$\angle AOB=\angle COD$ Vertically opposite angles)
$\Delta AOB\cong \Delta COD$ (SAS congruence rule)
AB=CD (By CPCT) …(1)
And $\angle OAB=\angle OCD$ (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 $\Delta AODand\Delta COD$
$\Delta AOD\cong \Delta COD$ (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 $\Delta ADCand\Delta BCD,$
AD=BC (Already proved)
AC=BD (Given)
DC=CD (Common)
$\Delta ADC\cong \Delta BCD$ (SSS Congruence rule)
$\angle ADC=\angle BCD$ (By CPCT)
However, $\angle ADC+\angle BCD={180}^{\circ }$ (Co-interior angles)
$\angle ADC+\angle ADC={180}^{\circ }$
$2\angle ADC={180}^{\circ }$
$\angle ADC={90}^{\circ }$
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 ${90}^{\circ }$.
Therefore, ABCD is a square.