Question

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

 

Answer 1

$$\text{ AE is the bisector of }\mathrm{\angle }A.$$
$$\therefore \mathrm{\angle }DAE=\mathrm{\angle }BAE=x$$
$$\mathrm{\angle }BAE=\mathrm{\angle }AED=x(\text{ alternate angles })$$
$$\text{ Since opposite angles in ∆ ADE are equal, ∆ ADE is an isosceles triangle }.$$
$$\therefore AD=DE=6cm(\text{ sides opposite to equal angles })$$
$$AB=CD=10cm$$
$$CD=DE+EC$$
$$\Rightarrow EC=CD-DE$$
$$\Rightarrow EC=10-6=4cm$$
$$\mathrm{\angle }DEA=\mathrm{\angle }CEF=x(\text{ vertically opposite angle })$$
$$\mathrm{\angle }EAD=\mathrm{\angle }EFC=x(\text{ alternate angles })$$
$$\text{ Since opposite angles in }∆EFC\text{ are equal, ∆ EFC is an isosceles triangle }.$$
$$\therefore CF=CE=4\text{ cm (sides opposite to equal angles })$$
$$\therefore CF=4\text{ cm }$$