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Question
In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F.
Find the length of CF.
Solution
Given AB = 20 cm and AD = 12 cm.
∠DAE = ∠BAF = x and
∠DAE = ∠AFB = x
In ΔABF
∠BAF = ∠BFA
Hence, ABF is an isosceles triangle
So AB = BF = 20
BF = 20
∠ADE = ∠ABF
∴ AD = BC = 12
BC + CF = 20
12 + CF = 20
CF = 20 - 12
CF = 8 cm
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