3. In a quadrilateral ABCD, equal diagonals AC andBD intersect at P, s

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

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

Answer»

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.