Question

In the following figure, $\angle BAC={60}^o$ and $\angle ABC={65}^o$.
Prove that :
$CF>AF$

Answer 1

$\angle BAC={60}^o$

$\angle ABC={65}^o$

$\therefore \angle ACB=180-60-65$

$\angle ACB={55}^o$

$\because AD$ is bisecting $\angle BAC$

$\therefore \angle FAC=300$

and $CE$ is bisecting $\angle ACB$

$\therefore \angle ACE=27.5$

$\because \angle FAC$ is angle opp. to $CF$

$\angle ACE$ is angle opp. to $AF$

$\because \angle FAC>\angle ACE$

$\therefore CF>AF$.