Need to find the length of a segment in a triangle

Question

I am currently stuck on a geometry problem involving a triangle and a parallel line. The problem is as follows:

Given $\triangle ABC$, where $AB = 7$ units. Let $l$ be a line parallel to side $AB$, intersecting sides $AC$ and $BC$ at points $P$ and $Q$, respectively. If $AP = 5$ units and $PC = 3$ units, what is the length of segment $PQ$?

I have attempted to solve this problem using the fact that $l$ is parallel to $AB$, which implies that $\triangle ABP$ and $\triangle CQP$ are similar. Using this similarity, I found that $CQ = \frac{3}{7}PQ$ and $BP = \frac{4}{7}PQ$. However, I am not sure how to proceed from here to find the length of $PQ$.

Answer is $PQ=\frac{21}{8}$ units.

I would greatly appreciate any hints or suggestions on how to solve this problem. Thank you in advance!

Answer 1

which implies that triangles $ABP$ and $CQP$ are similar.

This is incorrect, probably you have a typo here. It should be $\triangle CAB$ is similar to $\triangle CPQ$, then we have:

$$\frac{CP}{CA}=\frac{PQ}{AB}\Rightarrow \frac{3}{8}=\frac{PQ}{7}\Rightarrow PQ=\frac{21}{8}$$

Answer 2

Please see the diagram below:

First of all, triangles ABP and CQP are not similar. With this step, you have got an incorrect answer.

You should prove the similarity between triangles ABC and PQC.

Once this is done,

$$\frac{AB}{AC} = \frac{PQ}{PC}$$ Since, AC = AP + PC you can use the above equation to find PQ.