In given figure, ${\text{DE}}\parallel {\text{AC}}$and ${\text{DC}}\parallel {\text{AP}}$, Prove that $\dfrac{{{\text{BE}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{BC}}}}{{{\text{CP}}}}$.

Hint:
 The side containing BE and EC are the parts of the triangle $\vartriangle ABC$. Similarly BC and CP are parts of $\vartriangle ABP$.
Following theorem can be useful in dealing with these types of questions.
Theorem 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In $\vartriangle PQR,{\text{ }}ST\parallel QR$,
Hence, by theorem, $\dfrac{{PS}}{{SQ}} = \dfrac{{PT}}{{TR}}$

Complete step by step solution:
Step 1: The given figure

Step 2: In figure 3: given that
${\text{DE}}\parallel {\text{AC}}$and ${\text{DC}}\parallel {\text{AP}}$,
To prove: $\dfrac{{{\text{BE}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{BC}}}}{{{\text{CP}}}}$.
Step 3
In $\vartriangle BAC$, a line parallel to side AC intersects other two sides BA and BC at D and E respectively.
$ \Rightarrow $ ${\text{DE}}\parallel {\text{AC}}$,
Hence, \[\dfrac{{BE}}{{EC}} = \dfrac{{BD}}{{DA}}\] (according to Theorem 1) …… (1)
Step 4
In $\vartriangle PBA$, a line parallel to side AP intersects other two sides BA and BP at D and C respectively.
$ \Rightarrow $ ${\text{DC}}\parallel {\text{AP}}$,
Hence, \[\dfrac{{BC}}{{CP}} = \dfrac{{BD}}{{DA}}\] (according to theorem 1) …… (2)
Step 5
From (1) and (2),
Hence, $\dfrac{{{\text{BE}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{BC}}}}{{{\text{CP}}}}$ proved.
Final answer: $\dfrac{{{\text{BE}}}}{{{\text{EC}}}}{\text{ = }}\dfrac{{{\text{BC}}}}{{{\text{CP}}}}$ has been proved.


Note:
The converse (or opposite) of the above mentioned theorem is also true.
i.e. if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Students must focus that in theorem1, the ratio of the parts of the sides of the triangle is taken, not the whole side of the triangle.
For example: in $\vartriangle BAC$,
Ratio $\dfrac{{{\text{BE}}}}{{{\text{BC}}}}$and ratio $\dfrac{{{\text{BD}}}}{{{\text{BA}}}}$ are invalid for theorem 1 stated in hint of the solution.