# In a parallelogram ABCD, AB = 10 cm, AD = 6 cm.

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Question:

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

Solution:

$\mathrm{AE}$ is the bisector of $\angle \mathrm{A}$.

$\therefore \angle \mathrm{DAE}=\angle \mathrm{BAE}=x$

$\angle \mathrm{BAE}=\angle \mathrm{AED}=x$ (alternate angles)

Since opposite angles in $\Delta \mathrm{ADE}$ are equal, $\Delta \mathrm{ADE}$ is an isosceles triangle.

$\therefore \mathrm{AD}=\mathrm{DE}=6 \mathrm{~cm}$ (sides opposite to equal angles)

$\mathrm{AB}=\mathrm{CD}=10 \mathrm{~cm}$

$\mathrm{CD}=\mathrm{DE}+\mathrm{EC}$

$\Rightarrow \mathrm{EC}=\mathrm{CD}-\mathrm{DE}$

$\Rightarrow \mathrm{EC}=10-6=4 \mathrm{~cm}$

$\angle \mathrm{DEA}=\angle \mathrm{CEF}=x$ (vertically opposite angle)

$\angle \mathrm{EAD}=\angle \mathrm{EFC}=x$ (alternate angles)

Since opposite angles in $\Delta \mathrm{EFC}$ are equal, $\Delta \mathrm{EFC}$ is an isosceles triangle.

$\therefore \mathrm{CF}=\mathrm{CE}=4 \mathrm{~cm}$ (sides opposite to equal angles)

$\therefore \mathrm{CF}=4 \mathrm{~cm}$