ABCD is a trapezium with AB parallel CD and angle BCD=twice angle DAB. If DC=a and BC=b, find the length of AB.
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Answer:
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Step-by-step explanation:
Given:
ABCD is a trapezium
AB // CD
∠BCD = 2∠DAB
DC = a
BC = b
To find:
The length of AB in terms of a and b
Construction:
Take a point E on AB and join C and E such that CE // DA as shown in the attached figure
Solution:
Let's assume, ∠DAB = y
∴ ∠BCD = 2∠DAB = 2y° .... (i)
We are given,
AB // CD
⇒ AE // CD
⇒ AECD is a parallelogram .... [from construction we have CE//DA]
∴ CD = AE = a ..... (ii) ..... [Opposite facing sides of a parallelogram are equal in length]
We know that → the sum of the angles of two adjacent sides of a trapezium is equal to 180°
∴ ∠B + ∠BCD = 180°
substituting the value of ∠BCD from (i)
⇒ ∠B = (180 - 2y)° ..... (iii)
Also,
∵ CE // DA
⇒ ∠CEB = ∠DAB ..... [alternate angles]
⇒ ∠CEB = ∠DAB = y° .... (iv)
In Δ CBE, using the angle sum property of a triangle, we get
∠ECB + ∠CEB + ∠B = 180°
substituting from (iii) & (iv) we get
⇒ ∠ECB + y° + (180 - 2y)° = 180°
⇒ ∠ECB + y° + 180° - 2y° = 180°
⇒ ∠ECB + y° - 2y° = 0
⇒ ∠ECB - y° = 0
⇒ ∠ECB = y° ..... (v)
From (iv) & (v), we get
∠CEB = ∠ECB = y°
∴ BC = BE ...... [sides opposite to equal angles are also equal to each other]
⇒ BC = BE = b ..... (vi) ..... [∵ BE = b (given)]
Now, from the figure, we get
AB = AE + BE
substituting the value of AE & BE from (ii) & (vi), we get
⇒ AB = a + b
