Question

Diagonals of a square are perpendicular bisectors of each other ​

Answer 1

Answer:

The diagonals of a square are perpendicular bisectors of one another. As a result, Their intersection forms four right angles, and each diagonal is split into two congruent pieces. Therefore, if given the length of a diagonal, the length of one segment of that diagonal is half of the length of the entire diagonal.

Step-by-step explanation:

Answer 2

Given: The quadrilateral is a square.

To prove: Diagonals of a square are equal and bisect each other at right angles. AC = BD, OA = OC, OB = OD, and ∠AOB = 90°

Proof:

Let ABCD be a square,

Let the diagonals AC and BD intersect each other at a point O.

In ΔABC and ΔDCB,

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

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

BC = CB (Common side)

∴ ΔABC ≅ ΔDCB (By SAS congruence rule)

∴ AC = DB (By CPCT) ------------ (1)

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) ------------ (2)

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) --------- (3)

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

2∠AOB = 180° [From equation (3)]

∠AOB = 90° ------------- (4)

Hence, from equations (1), (2) and (4), we see that the diagonals of a square are equal and bisect each other at right angles.

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