show that the diagonals of a square are equal and perpendicular to each other
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5
Sol:
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.
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.
Answered by
11
Given :- ABCD is a square.
To proof :- AC = BD and AC ⊥ BD
Proof :- In △ ADB and △ BCA
AD = BC [ Sides of a square are equal ]
∠BAD = ∠ABC [ 90° each ]
AB = BA [ Common side ]
△ADB ≅ △BCA [ SAS congruency rule ]
⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]
In △AOB and △AOD
OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]
AB = AD [ Sides of a square are equal ]
AO = AO [ Common side ]
△AOB ≅ △ AOD [ SSS congruency rule ]
⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]
∠AOB + ∠AOD = 180° [ Linear pair ]
∠ AOB = ∠AOD = 90°
⇒ AO ⊥ BD
⇒ AC ⊥ BD
Hence proved, AC = BD and AC ⊥BD
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