In $\triangle A B C, A B=8, B C=11$, and $A C=6$. The points $P$ and $Q$ are on $B C$ such that $\triangle P B A$ and $\triangle Q A C$ are each similar to $\triangle A B C$. What is the length of $P Q ?$

![](https://cdn.mathpix.com/cropped/2024_04_17_d13f9eaf1d614231018ag-3.jpg?height=360&width=534&top_left_y=300&top_left_x=855)
From the similarity of $\triangle P B A$ and $\triangle A B C$, we get $\frac{P B}{A B}=\frac{A B}{C B}$.

Substituting, we get $\frac{P B}{8}=\frac{8}{11}$ which can be rearranged to get $P B=\frac{64}{11}$.

From the similarity of $\triangle Q A C$ and $\triangle A B C$, we get $\frac{Q C}{A C}=\frac{A C}{B C}$.

Substituting, we get $\frac{Q C}{6}=\frac{6}{11}$ which can be rearranged to get $Q C=\frac{36}{11}$.

We now know the lengths of $P B, Q C$, and $B C$. Since $P Q=B C-P B-Q C$, we get

$$
P Q=11-\frac{64}{11}-\frac{36}{11}=\frac{121-64-36}{11}=\frac{21}{11}
$$

ANSWER: $\frac{21}{11}$