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

\( l \| \mathrm{m} \) and line segments \( \mathrm{AB}, \mathrm{CD} \) and \( \mathrm{EF} \) are concurrent at point \( \mathrm{P} \).

To do:

We have to prove that \( \frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}} \).

Solution:

In $\triangle A P C$ and $\triangle B P D$

$\angle A P C=\angle B P D$             (Vertically opposite angles)

$\angle P A C=\angle P B D$                  (Alternate angles)

Therefore, by AA similarity,

$\triangle A P C \sim \triangle B P D$

This implies,

$\frac{A P}{P B}=\frac{A C}{B D}=\frac{P C}{P D}$..........(i)

In $\triangle A P E$ and $\triangle B P F$

$\angle A P E=\angle B P F$          (Vertically opposite angles)

$\angle P A E=\angle P B F$            (Alternate angles)

Therefore, by AA similarity,

$\triangle A P E  \sim \triangle B P F$

This implies,

$\frac{A P}{P B}=\frac{A E}{B F}=\frac{P E}{P F}$........(ii)

In $\triangle P E C$ and $\triangle P F D$

$\angle E P C=\angle F P D$              (Vertically opposite angles)

$\angle P C E =\angle P D F$                (Alternate angles)

Therefore, by AA similarity,

$\triangle P E C \sim \triangle P F D$

This implies,

$\frac{P E}{P F} =\frac{P C}{P D}=\frac{E C}{F D}$........(iii)

From (i), (ii) and (iii), we get,

$\frac{A P}{P B}=\frac{A C}{B D}=\frac{A E}{B F}=\frac{P E}{P F}=\frac{E C}{F D}$

$\frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}$

Hence proved.