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