Question

show that the diagonals of a square are equal and bisect each other at right​

Answer 1

Given: ABCD is a square. the diagonals AC and BD intersect at O.

TPT: AC=BD and bisect each other at 90°

Proof: in the triangle ADC and triangle BCD,

AD=BC,

ADC=BCD=90°

and CD is common,

therefore by SAS congruency triangle ADC and triangle BCD are congruent triangle.

hence AC=BD, i.e. diagonals are of equal length.

and CDB=ACD i.e. CDO=OCD

in the triangle OCD, OD=OC [sides opposite to equal angles are equal in length]

similarly we can show that OC=OB

Hence by the symmetry OD=OB=OC=OA

By SSS congruency triangle COD and triangle COB will be congruent triangles. [since CD=CB, OD=OB and OC is common]

hence COD=COB=180/2=90°

hence diagonals bisect at right angle.....✌️❣️

Answer 2

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)

So, ΔABC ≅ ΔDCB (By SAS congruency)

Hence, 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)

So, ΔAOB ≅ ΔCOD (By AAS congruence rule)

Hence, 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)

So, ΔAOB ≅ ΔCOB (By SSS congruency)

Hence, ∠AOB = ∠COB (By CPCT)

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

2∠AOB = 180°

∠AOB = 90°

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