Question

In a parallelogram $ABCD$, $AB=10\mathrm{cm}$ and $AD=6\mathrm{cm}$ . The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$. Find the length of $CF$.

Answer 1

Solve for the length of $CF$

Given, $AB=10\mathrm{cm}$ and $AD=6\mathrm{cm}$

$\angle DAE=\angle EFC$ $[\because$Alternate angles$]$

$\angle BAE=\angle DAE$ $[\because AF$ bisects $\angle A]$

$\Rightarrow \angle BAE=\angle DAE$

$\Rightarrow AB=BF$ $[\because$Sides opposite to equal angles are equal$]$

$\Rightarrow BF=10\mathrm{cm}$

$AD=BC$ $[\because$Opposite sides of a parallelogram are equal and parallel$]$

$\begin{array}{cc}\Rightarrow & BC+CF=10\mathrm{cm}\\ \Rightarrow & CF=10-6\\ \Rightarrow & CF=4\mathrm{cm}\end{array}$

Hence $\mathrm{CF}=4\mathrm{cm}$.