Question

If abcd is a square then prove that diagonals are equal and bisects each other at right angles

Answer 1

SOLUTION :-

Consider ΔADC and ΔBDC

 ∠ BCD = ∠ ADC = 90° ( angles in a square )

 BC = BD ( Sides of the square )

 DC = DC ( Common side )

 ∴ Δ ADC ≅ Δ BDC

 So, BD = AC ( by CPCT )      

Hence diagonals are equal .

 

Consider ΔAOB and ΔDOC

 ∠ AOB = ∠ DOC ( vertically opposite angles )

 ∠ ABO = ∠ ODC ( alternate interior angles )

 ∠ BAD = ∠ OCD ( alternate interior angles )

  ∴ ΔAOB ≅ ΔDOC

  So, OA = OC and OB = OD ( by CPCT )                        .

 Hence diagonals bisect each other.

 Consider ΔAOB and ΔAOD

  OB = OD ( diagonals bisect each other )

  AB = AD ( sides of the square )

  AO = AO ( common side )

  ∴ ΔAOB ≅ ΔAOD

   So, ∠ AOB = ∠ AOD  

We know that,

 ∠ AOB + ∠ AOD = 180  ( linear pair )

  ∴  ∠ AOB = ∠ AOD = 90°

So diagonals 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.