Question

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

Answer 1

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

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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.

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Answer 2

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

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${\blue{\sf\underline{Given}}}$

A square ABCD whose diagonals AC and BC intersect at O.

${\green{\sf\underline{To\:prove}}}$

(i) AC = BD

(ii) OA = OC and OB = OD

(iii) AC ⊥ BD.

${\blue{\sf\underline{Proof}}}$

(i) In ∆ABC and BAD,we have

AB = BA (commom)

BC = AD (sides of a square)

∠ABC = ∠BAD (each equal to 90°)

∴ ∆ABC ≅ ∆BAD (SAS-criterion).

And so, AC = BD (c.p.c.t).

(ii) Now,ABCD is a square and therefore a ||gm.

And so, OA = OC and OB = OD

[∴ diagonals of a ||gm bisect each other].

(iii) Now,in ∆AOB and AOD, we have

OB = OD [∴diagonals of a ||gm bisect each other]

AB = AD (sides of a square)

AO = AO (common)

∴ ∆AOB ≅ ∆AOD (SSS-criterion)

∴ ∠AOB = ∠AOD.

But,∠AOB + ∠AOD = 180° (linear pair)

∴ ∠AOB = ∠AOD = 90°

Thus, AO ⊥ BD, i.e., AC ⊥ BD.

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