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Question
In the following figure, AE and BC are equal and parallel and the three sides AB, CD, and DE are equal to one another. If angle A is 102o. Find angles AEC and BCD.
Solution
In the given figure
Given that AE = BC
We have to find ∠AEC and ∠BCD
In the quadrilateral AECB
(AE = BC and AE || BC)
So quadrilateral is a parallelogram
⇒ ∠BAE + ∠AEC = 180°
⇒ 102° + ∠AEC = 180°
⇒ ∠AEC = 78°
Also, ∠BAE = ∠BCE = 102°
AB = EC (because AECB is a parallelogram )
Now consider ΔECD
EC = ED = CD [Since AB = EC ]
Therefore ΔEDC is an equilateral triangle.
⇒ ∠ECD = 60°
∠BCD = ∠BCE + ∠ECD
⇒ ∠BCD = 102° + 60°
⇒ ∠BCD = 162°
Therefore ∠AEC = 78° and ∠BCD = 162°
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