In the figure,$l \parallel m$ and line segments $AB, CD$ and $EF$ are concurrent at point $P$. Prove that $\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$.

(N/A) Given $l \parallel m$ and line segments $AB, CD$ and $EF$ are concurrent at point $P$.
To prove: $\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$.
Proof:
In $\triangle APC$ and $\triangle BPD$,
$\angle APC = \angle BPD$ [Vertically opposite angles]
$\angle PAC = \angle PBD$ [Alternate interior angles,as $l \parallel m$]
Therefore,$\triangle APC \sim \triangle BPD$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AC}{BD} = \frac{PC}{PD}$ ......$(i)$
In $\triangle APE$ and $\triangle BPF$,
$\angle APE = \angle BPF$ [Vertically opposite angles]
$\angle PAE = \angle PBF$ [Alternate interior angles]
Therefore,$\triangle APE \sim \triangle BPF$ [By $AA$ similarity criterion]
Then,$\frac{AP}{PB} = \frac{AE}{BF} = \frac{PE}{PF}$ ......$(ii)$
In $\triangle PEC$ and $\triangle PFD$,
$\angle EPC = \angle FPD$ [Vertically opposite angles]
$\angle PCE = \angle PDF$ [Alternate interior angles]
Therefore,$\triangle PEC \sim \triangle PFD$ [By $AA$ similarity criterion]
Then,$\frac{PE}{PF} = \frac{PC}{PD} = \frac{CE}{FD}$ ......$(iii)$
From equations $(i), (ii)$ and $(iii)$,
$\frac{AP}{PB} = \frac{AC}{BD} = \frac{AE}{BF} = \frac{PE}{PF} = \frac{CE}{FD}$
Thus,$\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}$. Hence proved.