$\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$.
Question.
$\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$
$\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$
Solution:
In $\Delta \mathrm{ABE}$ and $\Delta \mathrm{CFB}$,
$\angle \mathrm{EAB}=\angle \mathrm{BCF}$ (opp. angles of parallelogram)
$\angle \mathrm{AEB}=\angle \mathrm{CBF}$ (Alternate interior angles, $\mathrm{As} \mathrm{AE} \| \mathrm{BC}$ )
$\therefore \quad$ By AA similarity
$\Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$
In $\Delta \mathrm{ABE}$ and $\Delta \mathrm{CFB}$,
$\angle \mathrm{EAB}=\angle \mathrm{BCF}$ (opp. angles of parallelogram)
$\angle \mathrm{AEB}=\angle \mathrm{CBF}$ (Alternate interior angles, $\mathrm{As} \mathrm{AE} \| \mathrm{BC}$ )
$\therefore \quad$ By AA similarity
$\Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$