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

#### Solution

\[\text{ AE is the bisector of } \angle A . \]

\[ \therefore \angle DAE = \angle BAE = x\]

\[ \angle BAE = \angle AED = x (\text{ alternate angles })\]

\[\text{ Since opposite angles in ∆ ADE are equal, ∆ ADE is an isosceles triangle } . \]

\[ \therefore AD = DE = 6 cm (\text{ sides opposite to equal angles })\]

\[AB = CD = 10 cm \]

\[CD = DE + EC\]

\[ \Rightarrow EC = CD - DE\]

\[ \Rightarrow EC = 10 - 6 = 4 cm\]

\[\angle DEA = \angle CEF = x (\text{ vertically opposite angle })\]

\[\angle EAD = \angle EFC = x (\text{ alternate angles })\]

\[\text{ Since opposite angles in } ∆ EFC \text{ are equal, ∆ EFC is an isosceles triangle } . \]

\[ \therefore CF = CE = 4 \text{ cm (sides opposite to equal angles })\]

\[ \therefore CF = 4\text{ cm }\]