  Question

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

Solution

Given that $$ABCD$$ is a square.To prove : $$AC = BD$$ and $$AC$$ and $$BD$$ bisect each other at right angles.Proof: (i)  In a $$Δ ABC$$ and $$Δ BAD$$,$$AB = AB$$ ( common line)$$BC = AD$$ ( opppsite sides of a square)$$∠ABC = ∠BAD$$ ( = 90° )$$Δ ABC ≅ Δ BAD$$( By SAS property)$$AC = BD$$ ( by CPCT).(ii) In a $$Δ OAD$$ and $$Δ OCB,$$$$AD = CB$$ ( 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. MathematicsRS AgarwalStandard IX

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