Question

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

Answer 1

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: $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$

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]

$\frac{AE}{BF}=\frac{AP}{BP}=\frac{EP}{FP}$ (I)

In $\Delta CEP$ and $\Delta DFP$,

$l||m$ [Given]

[ALternate interior angles] $\{\begin{array}{c}\angle 7=\angle 8\\ \angle 5=\angle 6\end{array}$

In $\Delta CEP$ and $\Delta DFP$, [By AA similarity criterion]

$\therefore \frac{CE}{DF}=\frac{CP}{DP}=\frac{EP}{FP}$ (II)

In $\Delta ACP$ and $\Delta BDP$,

$l||m$ [Given]

[Alternate interior angles] $\{\begin{array}{c}\angle 1=\angle 2\\ \angle 5=\angle 6\end{array}$

$\Delta ACP$ and $\Delta BDP$, [By AA similarity criterion]

$\frac{AC}{BD}=\frac{AP}{BP}=\frac{CP}{DP}$ (III)

$\frac{AP}{PB}=\frac{AC}{BD}=\frac{CP}{DP}=\frac{CE}{DF}=\frac{EP}{FP}=\frac{AE}{BF}$ [From (I),(II) and (III)]]

$\frac{AC}{BD}=\frac{AE}{BF}=\frac{CE}{DF}$

Hence proved,