Question

show that the diagonal of a square are equal and bisect each other at right please answer question

Answer 1

Show that the diagonals of a square are equal and bisect each other at right angles.

Sol: We have a square ABCD such that its diagonals AC and BD intersect at O.
(i) To prove that the diagonals are equal, i.e. AC = BD
In ΔABC and ΔBAD, we have
AB = BA
[Common]
BC = AD
[Opposite sides of the square ABCD]
∠ABC = ∠BAD
[All angles of a square are equal to 90°]
∴ΔABC ≌ ΔBAD
[SAS criteria]
⇒Their corresponding parts are equal.
⇒AC = BD
...(1)
(ii) To prove that 'O' is the mid-point of AC and BD.
∵ AD || BC and AC is a transversal.
[∵ Opposite sides of a square are parallel]
∴∠1 = ∠3
[Interior alternate angles]
Similarly, ∠2 = ∠4
[Interior alternate angles]
Now, in ΔOAD and ΔOCB, we have
AD = CB
[Opposite sides of the square ABCD]
∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved

ΔOAD ≌ ΔOCB [ASA criteria]

∴Their corresponding parts are equal.
⇒OA = OC and OD = OB
⇒O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O. ...(2)
(iii) To prove that AC 1 BD.
In ΔOBA and ΔODA, we have
OB = OD[Proved]

BA =DA[Opposite sides of the square]

OA = OA[Common]

∴ΔOBA ≌ ΔODA[SSS criteria]
⇒Their corresponding parts are equal.
⇒ ∠AOB = ∠AOD
But ∠AOB and ∠AOD form a linear pair.

∠AOB + ∠AOD = 180°

∠AOB = ∠AOD = 90°
⇒AC ⊥ BD
...(3)
From (1), (2) and (3), we get AC and BD are equal and 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.