Question

In a parallelogram $ABCD,AB=20cm$ and $AD=12cm$. The bisector of angle $A$ meets $DC$ at $E$ and $BC$ produced at $F$. Find the length of $CF$.

Answer 1

Here, $AB=20cm,AD=12cm$

$\therefore$ $DC=AB=20cm$

$\Rightarrow$ $AD=BC=12cm$

Bisector of $\angle A$ meets $DE$ on $E.$ Produced $AE$ and $BC$ to meet at point $F.$ Extend $AD$ to $G.$ From $F$ draw $HF\parallel CD.$

We have $CD\parallel FH$ and $CF\parallel DH.$

$\therefore$ $DCFH$ is a parallelogram.

Also, $AB\parallel FH$ and $AH\parallel BF$

$\therefore$ $ABFH$ is also parallelogram.

In $△AHF$ and $△ABF$

$\Rightarrow$ $\angle AHF=\angle ABF$ [ Opposite angles are equal ]

$\Rightarrow$ $AF=AF$ [ Common side ]

$\Rightarrow$ $\angle HAF=\angle FAB$ [ Since, $AF$ divides $\angle HAB$ ]

$\Rightarrow$ $△AFH\cong △ABF$ [ By AAS congruence ]

$\Rightarrow$ $AB=AH$ [ CPCT ]

$\Rightarrow$ $AB=AH=AD+DH=AD+CF$ [ Since, $DAFH$ is a parallelogram ]

$\therefore$ $CF=AB-AD=(20-12)cm=8cm$