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
Answers
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
∘
∴ ∠AOB = ∠AOD = 90
∘
∴ We have proved that diagonals of square are equal and perpendicular to each other.