Question

Show that diagonals of square are equal and bisects each other at 90°

Answer 1

Answer:                                        

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 = 1800       (Linear pair)

2∠AOB = 1800

∠AOB = 900

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

Answer 2

Step-by-step explanation:

Given that ABCD is a square.

To prove : AC=BD and AC and BD bisect each other at right angles.

Proof:

(i) In a ΔABC and ΔBAD,

AB=AB ( common line)

BC=AD ( opppsite sides of a square)

∠ABC=∠BAD ( = 90° )

ΔABC≅ΔBAD( By SAS property)

AC=BD ( by CPCT).

(ii) In a ΔOAD and ΔOCB,

AD=CB ( opposite sides of a square)

∠OAD=∠OCB ( transversal AC )

∠ODA=∠OBC ( transversal BD )

ΔOAD≅ΔOCB (ASA property)

OA=OC ---------(i)

Similarly OB=OD ----------(ii)

From (i) and (ii) AC and BD bisect each other.

Now in a ΔOBA and ΔODA,

OB=OD ( from (ii) )

BA=DA

OA=OA ( common line )

ΔAOB=ΔAOD----(iii) ( by CPCT

∠AOB+∠AOD=180° (linear pair)

2∠AOB=180°

∠AOB=∠AOD=90°

∴AC and BD bisect each other at right angles.