Question

Question 4
Show that the diagonals of a square are equal and bisect each other at right angles.

Answer 1

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)
$\angle ABO=\angle CDO$ (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.