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In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC are produced to meet at F. Find the length of CF.

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In ∆ADE and ∆CEF

∠1 = ∠3 …(i) (alternate interior angles)

as AD || BC (produced)

∠1 = ∠2 …(ii) (AF is the bisector of ∠A)

From (i) and (ii), we get

∠2 = ∠3

⇒ AB = BF (side opposite to equal angles are equal)

⇒ AB = BC + CF

⇒ 10 = 6 + CF

⇒ CF = 4 cm

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