# show that the diagonal of a square are equal and bisect each other prependicularly 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 AOB = 90º.

In ABC and DCB,

AB = DC (Sides of a square are equal to each other)

ABC = DCB (All interior angles are of 90 )

BC = CB (Common side)

ABC = DCB (By SAS congruency)

AC = DB (By CPCT)

Hence, the diagonals of a square are equal in length.

In AOB and COD,

AOB = COD (Vertically opposite angles)

ABO = CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal)

AOB = COD (By AAS congruence rule)

AO = CO and OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other.

In  AOB and COB,

As we had proved that diagonals bisect each other, therefore,

AO = CO

AB = CB (Sides of a square are equal)

BO = BO (Common)

AOB = COB (By SSS congruency)

AOB = COB (By CPCT)

However,AOB + COB = 180 (Linear pair)

2 AOB = 180º

AOB = 90º

Hence, the diagonals of a square bisect each other at right angles.

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You have a square ABCD.
Mid point is O.
Triangle ABC is congruent to triangle DCB by SAS congruence criterion. (AB=DC,Angle B=Angle C=90 degrees,BC=BC since its common)
Thus Ac is congruent to BD.(by CPCT),thus 1/2 Ac (i.e. OA) =1/2 BD (i.e. OB)
Now,in triangles AOb and BOC
AB=BC
(all sides equal in a square)
OA=OC(proved above)
and
OB=OB (common)
Thus by SSS congr. rule,the triangles are congruent and
by CPCT
Angle AOB=Angle BOC
The above angles form a linear pair
so,
Ang AOB+Ang.BOC=180 degrees
2 x Angle AOB=180 deg (Since aob=boc)
AOB=90 Degrees.Hence
BOC=90 deg too
And by vertically opposite angles property the other two angles are also right angles.

• 5 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 âˆ AOB = 90Âº.

In Î”ABC and Î”DCB,

AB = DC (Sides of a square are equal to each other)

âˆ ABC = âˆ DCB (All interior angles are of 90 )

BC = CB (Common side)

âˆ´ Î”ABC â‰… Î”DCB (By SAS congruency)

âˆ´ AC = DB (By CPCT)

Hence, the diagonals of a square are equal in length.

In Î”AOB and Î”COD,

âˆ AOB = âˆ COD (Vertically opposite angles)

âˆ ABO = âˆ CDO (Alternate interior angles)

AB = CD (Sides of a square are always equal)

âˆ´ Î”AOB â‰… Î”COD (By AAS congruence rule)

âˆ´ AO = CO and OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other.

In Î”AOB and Î”COB,

As we had proved that diagonals bisect each other, therefore,

AO = CO

AB = CB (Sides of a square are equal)

BO = BO (Common)

âˆ´ Î”AOB â‰… Î”COB (By SSS congruency)

âˆ´ âˆ AOB = âˆ COB (By CPCT)

However, âˆ AOB + âˆ COB = 180Âº (Linear pair)

2âˆ AOB = 180Âº

âˆ AOB = 90Âº

Hence, the diagonals of a square bisect each other at right angles.

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