Triangle Similarity and Parallel Lines

When a line segment ($\overline{PQ}$) is parallel to one side ($\overline{AB}$) of a triangle ($\Delta ABC$), and it intersects the other two sides ($\overline{BC}$ and $\overline{AC}$), it creates smaller similar triangles. In this case, $\overline{PQ} \parallel \overline{AB}$ implies that $\Delta CQP$ is similar to $\Delta CAB$.

Using Similarity Ratios

For similar triangles, the ratios of corresponding sides are equal. Therefore, we have: $ \frac{CQ}{CA} = \frac{CP}{CB} $

Calculating Side Ratios

We are given the ratio $CQ : QA = 1 : 3$. This means that for some value $x$, $CQ = 1x$ and $QA = 3x$. The total length of side $CA$ is the sum of $CQ$ and $QA$: $ CA = CQ + QA = 1x + 3x = 4x $ Now, we can find the ratio of $CQ$ to $CA$: $ \frac{CQ}{CA} = \frac{1x}{4x} = \frac{1}{4} $

Solving for BC Length

Substitute the known ratio $\frac{CQ}{CA} = \frac{1}{4}$ into the similarity equation: $ \frac{1}{4} = \frac{CP}{CB} $ We are given that $CP = 4$. Substituting this value: $ \frac{1}{4} = \frac{4}{CB} $ To find the length of $CB$ (which is the same as $BC$), we can cross-multiply: $ 1 \times CB = 4 \times 4 $ $ CB = 16 $ Thus, the length of $\overline{BC}$ is 16.