The diagonal BD of a parallelogram …

Given: The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC
To prove: DF {tex} \times{/tex} EF = FB {tex} \times{/tex} FA
 
Proof: In {tex}\triangle {/tex} FBE and {tex}\triangle {/tex} FDA,
{tex}\angle{/tex} FBE = {tex}\angle{/tex} FDA ........(1) .........[Alt.Int. {tex}\angle{/tex} s]
{tex}\angle{/tex} BFE = {tex}\angle{/tex} AFD (2)  ............[Vert. opp. {tex}\angle{/tex} s]
In view of (1) and (2),
{tex}\triangle {/tex}FBE ~ {tex}\triangle {/tex}FDA ..........AA similarity criterion
{tex}\therefore {/tex} {tex}\frac{{EF}}{{AF}} = \frac{{FB}}{{FD}}{/tex} .......... {tex}\because {/tex} Corresponding sides of two similar
{tex}\Rightarrow {/tex} {tex}\frac{{EF}}{{FA}} = \frac{{FB}}{{DF}}{/tex}
{tex}\Rightarrow {/tex} DF {tex} \times{/tex} EF = FB {tex} \times{/tex} FA