4. Show that the diagonals of a square are equal and bisect cach other at right angles.
Answers
Step-by-step explanation:
Given that ABCD is a square.
To prove : AC=BD and AC and BD bisect each other at right angles.
Proof:
(i) In a ΔABC and ΔBAD,
AB=AB ( common line)
BC=AD ( opppsite sides of a square)
∠ABC=∠BAD ( = 90° )
ΔABC≅ΔBAD( By SAS property)
AC=BD ( by CPCT).
(ii) In a ΔOAD and ΔOCB,
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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Given :
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- ABCD is a square
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- AB = BC = CD = DA
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- ∠A = ∠B = ∠C = ∠D = 90°
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To Prove :
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- AC = BD
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- ∠AOB=∠BOC=∠COD=∠DOA = 90°
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- OA= OB = OC = OD
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Proof :
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In △ACD and △BCD
AD = BC [ given ]
CD = CD [ common ]
∠D = ∠C [ equal equal to 90° ]
△ACD ≅ △BCD [ by SAS rule ]
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AC = BD [ by CPCT ]
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Since opposite sides are equal. Hence, sides are parallel.
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∠1 = ∠2 [ Alternate angles ]
∠3 = ∠4 [ Alternate angles ]
∠5 = ∠6 [ Alternate angles ]
∠7 = ∠8 [ Alternate angles ]
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In △AOB and △COD
AB = CD [ given ]
∠1 = ∠2 [ Alternate angles ]
∠3 = ∠4 [ Alternate angles ]
Therefore,△AOB ≅ △COD [ by ASA rule]
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OB = OD [ by CPCT ]
OC = OA [ by CPCT ]
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In △AOB and △BOC
AB = BC [ given ]
OB = OB [ common ]
OC = OA [ proved above ]
Therefore, △AOB ≅△BOC [by SSS rule ]
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∠AOB = ∠BOC [ by CPCT ]
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➨∠AOB + ∠BOC = 180° [ Linear Pair ]
Let ∠AOB and ∠BOC be x
➨ x + x = 180°
➨ 2x = 180°
➨ x = 180°/2
➨ x = 90°
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Hence, ∠AOB = ∠BOC = 90°
Similarly, ∠COD= ∠DOA = 90°.
