$\mathrm{E}$ 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 $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$

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

$\Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$
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