Given: $$l||m$$
Line segment $$AB,CD$$ and $$EF$$ are concurrent at $$P$$.
Points $$A,E$$ and $$C$$ are on line $$l$$.
Points $$D,F$$ and $$B$$ are on line $$m$$.
Refer image,
To prove: $$\dfrac { AE}{ BF} =\dfrac { AC }{ BD} =\dfrac { CE }{ FD } $$
Proof: In $$\Delta AEP$$ and $$\Delta BFP$$,
$$l||m$$ (Given)
$$\angle 1=\angle 2$$ []Alternate interior angles]
$$\angle 3=\angle 4$$ [same reason]
$$\therefore \Delta AEP\sim \Delta BFP$$, [By AA similarity criterion]
$$\dfrac { AE}{ BF} =\dfrac { AP }{ BP} =\dfrac { EP }{ FP } $$ (I)
In $$\Delta CEP$$ and $$\Delta DFP$$,
$$l||m$$ [Given]
[ALternate interior angles] $$\begin{cases} \angle 7 =\angle 8\\ \angle 5=\angle 6 \end{cases}$$
In $$\Delta CEP$$ and $$\Delta DFP$$, [By AA similarity criterion]
$$\therefore \dfrac { CE }{ DF} =\dfrac { CP}{ DP } =\dfrac { EP }{ FP } $$ (II)
In $$\Delta ACP$$ and $$\Delta BDP$$,
$$l||m$$ [Given]
[Alternate interior angles] $$\begin{cases} \angle 1 =\angle 2\\ \angle 5=\angle 6 \end{cases}$$
$$\Delta ACP$$ and $$\Delta BDP$$, [By AA similarity criterion]
$$\dfrac { AC}{ BD} =\dfrac { AP }{ BP} =\dfrac { CP }{ DP } $$ (III)
$$\dfrac { AP}{ PB} =\dfrac { AC }{ BD} =\dfrac { CP }{ DP } =\dfrac { CE}{ DF} =\dfrac { EP }{ FP} =\dfrac { AE }{ BF } $$ [From (I),(II) and (III)]]
$$\dfrac { AC}{ BD} =\dfrac { AE }{ BF} =\dfrac { CE }{ DF } $$
Hence proved,