Answers

2016-02-23T09:42:28+05:30
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.
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