Question

ABCD is a quadrilateral in which AB/CD and AD/BC. find angle b ,angle Came angle d​

Answer 1

Step-by-step explanation:

Here, AB∥CD [Given]

⇒ and AD∥EC [By construction]

∴ AECD is a parallelogram.

⇒ AD = EC [Opposite sides of parallelogram are equal]

⇒ But AD = EC [Given]

∴ EC = BC

∴ ∠CBE = ∠CEB ---- ( 1 )

⇒ ∠B + ∠CBE = 180

∘

[Linear pair] ---- ( 2 )

⇒ AD∥EC [By construction]

⇒ and transeversal AE intersects them

∴ ∠A + ∠CEB = 180

∘

---- ( 3 )

[Sum of adjacent angles of parallelogram is supplementary ]

⇒ ∠B + ∠CEB = 180

∘

[From ( 2 ) and ( 3 )]

⇒ But ∠CBE = ∠CEB [From ( 1 )]

∴ ∠A=∠B [Proved] -- ( 4 )

⇒ ∵ AB∥CD

⇒ ∠A + ∠D = 180

∘

[Sum Supplementary angles of parallelogram is 180 ]

⇒ and ∠B + ∠C = 180

∘

∴ ∠A + ∠D = ∠B + ∠C

⇒ But ∠A = ∠B [From ( 4 )]

∴ ∠C=∠D

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Answer 2

Step-by-step explanation:

Here, AB∥CD               [Given]

⇒  and AD∥EC            [By construction]

∴   AECD is a parallelogram.

⇒  AD = EC             [Opposite sides of parallelogram are equal]

⇒  But AD = EC       [Given]

∴   EC = BC

∴   ∠CBE = ∠CEB              ---- ( 1 )

⇒  ∠B + ∠CBE = 180∘  [Linear pair]      ---- ( 2 )

⇒  AD∥EC    [By construction]

⇒  and transeversal  AE intersects them

∴   ∠A + ∠CEB = 180∘     ---- ( 3 )  

[Sum of adjacent angles of parallelogram is supplementary ]

⇒  ∠B + ∠CEB = 180∘     [From ( 2 ) and ( 3 )]

⇒  But ∠CBE = ∠CEB       [From ( 1 )]

∴    ∠A=∠B     [Proved]    -- ( 4 )

⇒   ∵  AB∥CD

⇒   ∠A + ∠D = 180∘   [Sum 

Supplementary angles of parallelogram is 180∘]

⇒ and ∠B + ∠C = 180∘

∴    ∠A + ∠D = ∠B + ∠C

⇒  But ∠A = ∠B   [From ( 4 )]

∴   ∠C=∠D