Question

19. In the quadrilateral ABCD, AD is parallel to BC.K is a point on AB. Draw the line through A parallel to KC ineeting CD at Hand the line through B parallel to KD meeting CD at T. Find the ratio CD K HI​

Answer 1

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