Question

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

Answer 1

Given l || m and line segments AB, CD and EF are concurrent at point P

To prove: `(AE)/(BF) = (AC)/(BD) = (CE)/(FD)`

Proof: In ΔAPC and ΔBPD,

 ∠APC = ∠PBD   .....[Vertically opposite angles]

∠PAC = ∠PBD   .....[Alternative angles]

∴ ΔAPC ∼ ΔBPD   .....[By AA similarity criterion]

Then, `(AP)/(PB) = (AC)/(BD) = (PC)/(PD)`   ......(i)

In ΔAPE and ΔBPF,

∠APE = ∠BPF   ......[Vertically opposite angles]

∠PAE = ∠PBF   .....[Alternative angles]

∴ ΔAPE ∼ ΔBPF   .....[By AA similarity criterion]

Then, `(AP)/(PB) = (AE)/(BF) = (PE)/(PF)`   ......(ii)

In ΔPEC and ΔPFD,

∠EPC = ∠FPD   ......[Vertically opposite angles]

∠PCE = ∠PDF   .....[Alternative angles]

∴ ΔPEC ∼ ΔPFD  .....[By AA similarity criterion]

Then, `(PE)/(PF) = (PC)/(PD) = (EC)/(FD)`   ......(iii)

From equation (i), (ii) and (iii)

`(AP)/(PB) = (AC)/(BD) = (AE)/(BF) = (PE)/(PF) = (EC)/(FD)`

∴ `(AE)/(BF) = (AC)/(BD) = (CE)/(FD)`

Hence proved.