SHOW THAT THE DIAGONALS OF A SQUARE ARE EQUAL AND BISECT EACH EACH OTHER AT RIGHT ANGLES

Asked by DEVIKA BAHL | 5th Oct, 2010, 05:16: PM

Expert Answer:

Dear Student,

Consider a square ABCD whose diagonals AC and BD intersect each other at a point O.

We have to prove that AC = BD, OA = OC, OB = OD, and ∠AOB = 90º.
Consider ΔABC and ΔDCB,

AB = DC

∠ABC = ∠DCB = 90°

BC = CB (Common side)

∴ ΔABC ≅ ΔDCB (By SAS congruency)

∴ AC = DB (By CPCT)

Hence, the diagonals of the square ABCD are equal in length.

Now, consider ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠ABO = ∠CDO (Alternate interior angles)

AB = CD (Sides of a square are equal)

∴ ΔAOB ≅ ΔCOD (By AAS congruence rule)

∴ AO = CO and OB = OD (By CPCT)

Hence, the diagonals of the square bisect each other.

Similarly, in ΔAOB and ΔCOB,

AO = CO

AB = CB

BO = BO

∴ ΔAOB ≅ ΔCOB (By SSS congruency)

∴ ∠AOB = ∠COB (By CPCT)

∠AOB + ∠COB = 180º (Linear pair)

or, 2∠AOB = 180º

or, ∠AOB = 90º

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

 

We hope that clarifies your query.

Regards

Team

Topperlearning

Answered by  | 5th Oct, 2010, 07:17: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.