Question

show that the diagonals of a square are equal and bicect each other at right angles​

Answer 1

Let ABCD be a square such that its diagonals AC and BD intersect at O

(i) To prove that the diagonals are equal, we need to prove AC = BD.

In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

AC = BD [By C.P.C.T.] …(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]

∴ ∠1 = ∠3

[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved]

∴ ∆OAD ≅ ∆OCB [By ASA congruency]

⇒ OA = OC and OD = OB [By C.P.C.T.]

i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii) In ∆OBA and ∆ODA, we have

OB = OD [Proved]

BA = DA [Sides of a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency]

⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)

∵ ∠AOB and ∠AOD form a linear pair.

∴∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD …(4)

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.

(っ◔◡◔)っ ♥ itzcottoncandy65 ♥