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.

Answer 1

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