Question

3. In a quadrilateral ABCD, equal diagonals AC and
BD intersect at P, such that AP = PC and BP = PD,
also ZBPC = 90°, then quadrilateral is exactly
(a) a parallelogram (b) a square
a rhombus
(a rectangle​

Answer 1

Answer:

(b) square

Step-by-step 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

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∴ ∠AOB = ∠AOD = 90

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