Question

in the adjacent figure which area is parallel to BC and ab is equals to BAC is equal to Q BD is equal to AC is equals to S then show that p square minus Q square is equals to R square minus A square​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

⇒ In given figure ABCD is a square having all sides are equal and opposite sides parallel to each other. AC and BD are diagonals.

⇒ In △ABC and △BAD,

⇒ AB = AB [Common line]

⇒ BC = AD [Sides of square are equal.]

⇒ ∠ABC = ∠BAD [All four angles of square is 90

∘

]

⇒ △ABC ≅ △BAD [By SAS property]

⇒ In a △ OAD and △OCB,

⇒ AD = CB [sides of a square]

⇒ ∠OAD = ∠OCB [Alternate angle]

⇒ ∠ODA = ∠OBC [Alternate angle]

⇒ △OAD ≅ △OCB [By ASA Property]

⇒ So, OA = OC ----- ( 1 )

⇒ Similarly, OB = OD ------ ( 2 )

From ( 1 ) and ( 2 ) we get that AC and BD bisect each other.

⇒ Now, in △OBA and △ODA,

⇒ OB = OD [From ( 2 )]

⇒ BA = DA

⇒ OA = OA [Common line]

⇒ ∠AOB + ∠AOD ----- ( 3 ) [By CPCT]

⇒ ∠AOB + ∠AOD = 180

∘

[Linear pair]

⇒ 2∠AOB = 180

∘

∴ ∠AOB = ∠AOD = 90

∘

∴ We have proved that diagonals of square are equal and perpendicular to each other.

Answer 2

$\huge\underbrace\mathfrak \red{ANSWER }$

⇒ In given figure ABCD is a square having all sides are equal and opposite sides parallel to each other. AC and BD are diagonals.

⇒ In △ABC and △BAD,

⇒ AB = AB [Common line]

⇒ BC = AD [Sides of square are equal.]

⇒ ∠ABC = ∠BAD [All four angles of square is 90

∘

]

⇒ △ABC ≅ △BAD [By SAS property]

⇒ In a △ OAD and △OCB,

⇒ AD = CB [sides of a square]

⇒ ∠OAD = ∠OCB [Alternate angle]

⇒ ∠ODA = ∠OBC [Alternate angle]

⇒ △OAD ≅ △OCB [By ASA Property]

⇒ So, OA = OC ----- ( 1 )

⇒ Similarly, OB = OD ------ ( 2 )

From ( 1 ) and ( 2 ) we get that AC and BD bisect each other.

⇒ Now, in △OBA and △ODA,

⇒ OB = OD [From ( 2 )]

⇒ BA = DA

⇒ OA = OA [Common line]

⇒ ∠AOB + ∠AOD ----- ( 3 ) [By CPCT]

⇒ ∠AOB + ∠AOD = 180

∘

[Linear pair]

⇒ 2∠AOB = 180

∘

∴ ∠AOB = ∠AOD = 90

∘

∴ We have proved that diagonals of square are equal and perpendicular to each other.

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