Question

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

Answer 1

Given IIIm 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 $\Delta APC$ and $\Delta BPD$ $\angle APC=\angle BPD$ [vertically opposite angles]
$\angle PAC=\angle PBD$ [alternate angles]
$\therefore$ $\Delta APC\sim \Delta BPD$ [by AAA similarity criterion]
Then, $\frac{AP}{PB}=\frac{AC}{BD}=\frac{PC}{PD}$ ...(i)
In $\Delta APE$ and $\Delta BPD,\angle APE=\angle BPF$ [vertically opposite angles]
$\angle PAE=\angle PBF$ [ alternate angles]
$\Delta APE\sim \Delta BPF$ [by AAA similarity criterion]
Then, $\frac{AP}{PB}=\frac{AE}{BF}=\frac{PE}{PF}$ …..(ii)
In $\Delta PEC$ and $\Delta PFD$, $\angle EPC=\angle FPD$ [vertically opposite angles]
$\angle PCE=\angle PDF$ [ alternate angles]
$\therefore$ $\Delta PED\sim \Delta PFD$ [by AAA similarity criterion]
Then, $\frac{PE}{PF}=\frac{PC}{PD}=\frac{EC}{FD}$ ….(iii)
From, Eq.(i), (ii) and (iii),
$\frac{AP}{PB}=\frac{AC}{BD}=\frac{AE}{BF}=\frac{PE}{PF}=\frac{EC}{FD}$
$\therefore$ $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$