Question

The diagonals of a square are:

• A. equal and bisect each other but not at a right angle
• B. not equal and bisect each other at right angle.
• C. equal and bisect each other at a right angle
• D. equal and does not bisect each other but intersects at a right angle

Answer 1

The correct option is C

equal and bisect each other at a right angle

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC=BD, OA=OC,OB=OD, and $\angle AOBof{90}^{\circ })$

In $\Delta$ ABC and $\Delta$ DCB,
AB=DC (Sides of a square are equal to each other)
$\angle$ ABC = $\angle$ DCB (All interior angles are of ${90}^{\circ }$)
BC=CB (Common side)
$\therefore \Delta ABC\cong \Delta DCB$ (By SAS congruency)
$\therefore$ AC=DB (By CPCT)
Hence, the diagonals of a square are equal in length.

In $\Delta AOB$ and $\Delta COD$,
$\angle AOB=\angle COD$ ( Vertically opposite angles)
$angleABO=\angle CDO$ (Alternate interior angles)
AB=CD (sides of a square are always equal)
$\therefore \Delta AOB\cong \Delta COD$ (By AAS congruence rule)
$\Rightarrow$ AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.

In $\angle AOBand\angle COB,$
As we had proved that diagonals bisect each other, therefore,
AO=CO
AB=CB (Sides of a square are equal)
BO=BO (Common)
$\therefore \Delta AOB\cong \Delta COB$ (By SSS congruency)

$\angle$ AOB = $\angle$ COB (By CPCT)
However, $\angle AOB+\angle COB={180}^{\circ }$ (Linear pair)
$2\angle AOB={180}^{\circ }$
$\angle AOB={90}^{\circ }$
Hence , the diagonals of a square bisect each other at right angles.