In triangle $ABC$, points $E$ and $P$ lie on $AB$ such that $AE = EP = PB$. Points $F$ and $Q$ lie on $AC$ such that $AF = FQ = QC$. $EF$, $PQ$, and $BC$ are parallel straight lines. We need to verify that $BC = EF + PQ$.

1. Problem Description

In triangle $ABC$, points $E$ and $P$ lie on $AB$ such that $AE = EP = PB$. Points $F$ and $Q$ lie on $AC$ such that $AF = FQ = QC$. $EF$, $PQ$, and $BC$ are parallel straight lines. We need to verify that $BC = EF + PQ$.

2. Solution Steps

Since $AE = EP = PB$, we have $AE = \frac{1}{3}AB$ and $AP = \frac{2}{3}AB$.

Similarly, since $AF = FQ = QC$, we have $AF = \frac{1}{3}AC$ and $AQ = \frac{2}{3}AC$.

Since $EF$ is parallel to $BC$, triangle $AEF$ is similar to triangle $ABC$ (by the AA similarity criterion, as $\angle AEF = \angle ABC$ and $\angle AFE = \angle ACB$).

Therefore, the ratio of corresponding sides is equal:

$\frac{AE}{AB} = \frac{AF}{AC} = \frac{EF}{BC}$.

Since $AE = \frac{1}{3}AB$, we have $\frac{AE}{AB} = \frac{1}{3}$. Therefore, $\frac{EF}{BC} = \frac{1}{3}$, which implies

$EF = \frac{1}{3}BC$.

Similarly, since $PQ$ is parallel to $BC$, triangle $APQ$ is similar to triangle $ABC$.

Therefore, $\frac{AP}{AB} = \frac{AQ}{AC} = \frac{PQ}{BC}$.

Since $AP = \frac{2}{3}AB$, we have $\frac{AP}{AB} = \frac{2}{3}$. Therefore, $\frac{PQ}{BC} = \frac{2}{3}$, which implies

$PQ = \frac{2}{3}BC$.

Now we can compute $EF + PQ$:

$EF + PQ = \frac{1}{3}BC + \frac{2}{3}BC = \frac{1+2}{3}BC = \frac{3}{3}BC = BC$.

Therefore, $BC = EF + PQ$.

3. Final Answer

$BC = EF + PQ$ is verified.