1. **Problem statement:** Given triangle ABC with BC parallel to DE, and segments BD = 2x - 4, DC = 3x + 6, BE = x + 5, and EA = 8x - 5, find the value of $x$. 2. **Formula and rule:** Since BC is parallel to DE, by the Basic Proportionality Theorem (Thales' theorem), the segments on the transversal are proportional: $$\frac{BD}{DC} = \frac{BE}{EA}$$ 3. **Set up the proportion:** $$\frac{2x - 4}{3x + 6} = \frac{x + 5}{8x - 5}$$ 4. **Cross multiply:** $$(2x - 4)(8x - 5) = (3x + 6)(x + 5)$$ 5. **Expand both sides:** $$16x^2 - 10x - 32x + 20 = 3x^2 + 15x + 6x + 30$$ Simplify: $$16x^2 - 42x + 20 = 3x^2 + 21x + 30$$ 6. **Bring all terms to one side:** $$16x^2 - 42x + 20 - 3x^2 - 21x - 30 = 0$$ $$13x^2 - 63x - 10 = 0$$ 7. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=13$, $b=-63$, $c=-10$. Calculate discriminant: $$\Delta = (-63)^2 - 4 \times 13 \times (-10) = 3969 + 520 = 4489$$ Calculate square root: $$\sqrt{4489} = 67$$ Calculate roots: $$x = \frac{63 \pm 67}{26}$$ 8. **Find two possible values:** $$x_1 = \frac{63 + 67}{26} = \frac{130}{26} = 5$$ $$x_2 = \frac{63 - 67}{26} = \frac{-4}{26} = -\frac{2}{13}$$ 9. **Check for valid solution:** Since segment lengths must be positive, test $x=5$: - $2x - 4 = 2(5) - 4 = 6 > 0$ - $3x + 6 = 15 + 6 = 21 > 0$ - $x + 5 = 10 > 0$ - $8x - 5 = 40 - 5 = 35 > 0$ For $x = -\frac{2}{13}$, some segments become negative, so discard. **Final answer:** $$\boxed{x = 5}$$