In triangle $ABC$, points $E$ and $P$ are on side $AB$ such that $AE = EP = PB$. Points $F$ and $Q$ are on side $AC$ such that $AF = FQ = QC$. Lines $EF$, $PQ$, and $BC$ are parallel. The problem asks us to verify that $BC = EF + PQ$.

1. Problem Description

In triangle $ABC$, points $E$ and $P$ are on side $AB$ such that $AE = EP = PB$. Points $F$ and $Q$ are on side $AC$ such that $AF = FQ = QC$. Lines $EF$, $PQ$, and $BC$ are parallel. The problem asks us to verify that $BC = EF + PQ$.

2. Solution Steps

Since $AE = EP = PB$, we have $AE = \frac{1}{3} AB$ and $AP = \frac{2}{3} AB$. Also, since $AF = FQ = QC$, we have $AF = \frac{1}{3} AC$ and $AQ = \frac{2}{3} AC$.

Since $EF \parallel BC$, by the triangle proportionality theorem (also known as Thales' theorem or the side-splitter theorem), we have $\frac{AE}{AB} = \frac{AF}{AC}$. This is true since $\frac{AE}{AB} = \frac{\frac{1}{3}AB}{AB} = \frac{1}{3}$ and $\frac{AF}{AC} = \frac{\frac{1}{3}AC}{AC} = \frac{1}{3}$. Similarly, since $PQ \parallel BC$, we have $\frac{AP}{AB} = \frac{AQ}{AC}$. This is true since $\frac{AP}{AB} = \frac{\frac{2}{3}AB}{AB} = \frac{2}{3}$ and $\frac{AQ}{AC} = \frac{\frac{2}{3}AC}{AC} = \frac{2}{3}$.

Using the fact that $EF \parallel BC$, we can say that $\triangle AEF \sim \triangle ABC$. Then,

$\frac{EF}{BC} = \frac{AE}{AB} = \frac{AF}{AC} = \frac{1}{3}$, so $EF = \frac{1}{3} BC$.

Using the fact that $PQ \parallel BC$, we can say that $\triangle APQ \sim \triangle ABC$. Then,

$\frac{PQ}{BC} = \frac{AP}{AB} = \frac{AQ}{AC} = \frac{2}{3}$, so $PQ = \frac{2}{3} BC$.

We are asked to verify that $BC = EF + PQ$. Substituting the expressions for $EF$ and $PQ$ we found above, we have:

$BC = \frac{1}{3} BC + \frac{2}{3} BC$

$BC = \frac{3}{3} BC$

$BC = BC$

Therefore, the statement $BC = EF + PQ$ is true.

3. Final Answer

$BC = EF + PQ$ is verified.