Question

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

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

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]

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

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

$$l||m$$ [Given]

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

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

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

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

$$l||m$$ [Given]

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

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

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

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

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

Hence proved,