1. **State the problem:** In triangle $\triangle ABC$, a line parallel to side $BC$ meets $AB$ and $AC$ at points $P$ and $Q$, respectively. Given that $AP = QC$, $AB = 12$ cm, and $AQ = 2$ cm, find the length of $CQ$. 2. **Understand the setup:** Since $PQ \parallel BC$, triangles $\triangle APQ$ and $\triangle ABC$ are similar by the Basic Proportionality Theorem (also known as Thales' theorem). 3. **Write down proportionalities from similarity:** From similarity: $$\frac{AP}{AB} = \frac{AQ}{AC} = \frac{PQ}{BC}$$ 4. **Assign variables:** Since $AP = QC$, let: $$AP = QC = x$$ Given: $$AB = 12, \quad AQ = 2$$ We want to find $CQ$. 5. **Express the segments on $AB$ and $AC$:** On side $AB$: $$AB = AP + PB = x + PB = 12 \implies PB = 12 - x$$ On side $AC$: $$AC = AQ + QC = 2 + x$$ 6. **Apply similarity ratio:** From the similarity: $$\frac{AP}{AB} = \frac{AQ}{AC} \implies \frac{x}{12} = \frac{2}{2 + x}$$ 7. **Solve for $x$:** Cross-multiplied: $$x(2 + x) = 12 \times 2 \implies 2x + x^2 = 24$$ Rewrite: $$x^2 + 2x - 24 = 0$$ 8. **Solve quadratic equation:** Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=2$, $c=-24$: $$x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24)}}{2} = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm \sqrt{100}}{2}$$ $$x = \frac{-2 \pm 10}{2}$$ Two solutions: $$x = \frac{-2 + 10}{2} = 4 \quad \text{or} \quad x = \frac{-2 - 10}{2} = -6$$ Since length can’t be negative, $x = 4$ cm. 9. **Final answer:** $$CQ = x = 4 \text{ cm}$$