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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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View Full Answer

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 ∠AOB = 90º.

In ΔABC and ΔDCB,

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

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

BC = CB (Common side)

∴ ΔABC ≅ ΔDCB (By SAS congruency)

∴ AC = DB (By CPCT)

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

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

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

∴ ΔAOB ≅ ΔCOD (By AAS congruence rule)

∴ AO = CO and OB = OD (By CPCT)

Hence, the diagonals of a square bisect each other.

In ΔAOB and ΔCOB,

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

AO = CO

AB = CB (Sides of a square are equal)

BO = BO (Common)

∴ ΔAOB ≅ ΔCOB (By SSS congruency)

∴ ∠AOB = ∠COB (By CPCT)

However, ∠AOB + ∠COB = 180º (Linear pair)

2∠AOB = 180º

∠AOB = 90º

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

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