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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.


490294_463877_ans_a1d9648ef0814ec3ae10286550c4fa82.png

Mathematics
RS Agarwal
Standard IX

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