If ABCD is a quadrilateral in which AB|| CD and AD=BC, prove that. (i) <A=<B. (ii) AC=BD
Answers
Answer:
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
Step-by-step explanation:
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