In given figure,l`abs()`m and liner segments AB, CD and EF are concurrent at point P. Prove that `(AE)/(BF)=(AC)/(BD)=(CE)/(FD)`

Explanation

Given l`abs()`m and line segments AB, CD and EF are concurrent at point P
To prove `(AE)/(BF)=(AC)/(BD)=(CE)/(FD)`
Proof In `DeltaAPC and DeltaBPD, angleAPC=angleBPD` [vertically opposite angle]
`anglePAC=anglePBD` [alternate angle]
`therefore DeltaAPC~DeltaBPD` [by AAA similarity criterion]
Then, `(AP)/(PB)=(AC)/(BD)=(PC)/(PD)`
In `DeltaAPE` and `DeltaBPF, angleAPE=angleBPF` [vertically opposite angles]
`anglePAE=anglePBF` [alternate angles]
`therefore DeltaAPE~DeltaBPF` [by AAA similarity criterion]
Then, `(AP)/(PB)=(AE)/(BF)=(PE)/(PF)`....(ii)
In `DeltaPEC and DeltaPFD, angleEPC=angleFPD` [vertically opposite angle]
`anglePCE=anglePDF` [alternate angles]
`therefore DeltaPEC~DeltaPFD` [by AAA similarity criterion]
Then, `(PE)/(PF)=(PC)/(PD)=(EC)/(FD)`....(iii)
From Eqs. (i) ,(ii) and (iii),
`(AP)/(PB)=(AC)/(BD)=(AE)/(BF)=(PE)/(PF)=(EC)/(FD)`
`therefore (AE)/(BF)=(AC)/(BD)=(CE)/(FD)` Hence proved