1. **State the problem:** We have a right triangle ABC with a right angle at D on segment AC. Given lengths are $AC=20$ and $CD=12$. We need to find the length $BC$. 2. **Understand the setup:** Since $D$ is the foot of the perpendicular from $B$ to $AC$, triangle $BCD$ is a right triangle with right angle at $D$. 3. **Find length $AD$:** Since $AC=20$ and $CD=12$, then $$AD = AC - CD = 20 - 12 = 8.$$ 4. **Use the right triangle properties:** Triangle $BCD$ is right angled at $D$, so by the Pythagorean theorem, $$BC^2 = BD^2 + CD^2.$$ 5. **Use similar triangles:** Triangles $ABD$ and $BCD$ are similar (both right angled and share angle at $B$). Using similarity, $$\frac{AB}{BD} = \frac{BD}{BC}.$$ 6. **Express $BD$ in terms of $AB$ and $BC$:** From the similarity, $$BD^2 = AB \times BC.$$ 7. **Use Pythagorean theorem in triangle $ABD$:** $$AB^2 = AD^2 + BD^2 = 8^2 + BD^2 = 64 + BD^2.$$ 8. **Substitute $BD^2$ from step 6:** $$AB^2 = 64 + AB \times BC.$$ 9. **Use Pythagorean theorem in triangle $BCD$:** $$BC^2 = BD^2 + CD^2 = BD^2 + 12^2 = BD^2 + 144.$$ 10. **Substitute $BD^2$ from step 6:** $$BC^2 = AB \times BC + 144.$$ 11. **Let $AB = x$ and $BC = y$, then from step 8:** $$x^2 = 64 + xy,$$ from step 10: $$y^2 = xy + 144.$$ 12. **Rearrange both equations:** $$x^2 - xy - 64 = 0,$$ $$y^2 - xy - 144 = 0.$$ 13. **Subtract the first from the second:** $$(y^2 - xy - 144) - (x^2 - xy - 64) = 0,$$ which simplifies to $$y^2 - x^2 - 80 = 0,$$ so $$y^2 - x^2 = 80.$$ 14. **Factor difference of squares:** $$(y - x)(y + x) = 80.$$ 15. **From step 11, express $xy$ from first equation:** $$xy = x^2 - 64.$$ 16. **Try to find $x$ and $y$ by substitution or guess:** Since $x$ and $y$ are lengths, positive numbers. 17. **Solve system:** From step 13, $$y^2 = x^2 + 80,$$ substitute into second equation from step 11: $$y^2 - xy - 144 = 0,$$ replace $y^2$: $$(x^2 + 80) - xy - 144 = 0,$$ which simplifies to $$x^2 - xy - 64 = 0,$$ which is the same as the first equation. 18. **Solve quadratic for $y$ in terms of $x$:** From first equation, $$x^2 - xy - 64 = 0 \implies xy = x^2 - 64 \implies y = \frac{x^2 - 64}{x} = x - \frac{64}{x}.$$ 19. **Substitute $y$ into $y^2 = x^2 + 80$:** $$\left(x - \frac{64}{x}\right)^2 = x^2 + 80,$$ expand left side: $$x^2 - 2 \times x \times \frac{64}{x} + \frac{64^2}{x^2} = x^2 + 80,$$ which simplifies to $$x^2 - 128 + \frac{4096}{x^2} = x^2 + 80,$$ subtract $x^2$ from both sides: $$-128 + \frac{4096}{x^2} = 80,$$ add 128 to both sides: $$\frac{4096}{x^2} = 208,$$ so $$x^2 = \frac{4096}{208} = 19.6923,$$ then $$x = \sqrt{19.6923} \approx 4.44.$$ 20. **Find $y$:** $$y = x - \frac{64}{x} = 4.44 - \frac{64}{4.44} = 4.44 - 14.41 = -9.97,$$ which is negative and not possible for length. 21. **Try the other root for $x^2$:** Since quadratic in step 19 is $$x^4 - 208 x^2 + 4096 = 0,$$ solve for $x^2$: $$x^2 = \frac{208 \pm \sqrt{208^2 - 4 \times 4096}}{2} = \frac{208 \pm \sqrt{43264}}{2}.$$ Calculate: $$\sqrt{43264} \approx 208.01,$$ so $$x^2 = \frac{208 \pm 208.01}{2}.$$ Two solutions: $$x^2_1 = \frac{208 + 208.01}{2} = 208.005,$$ $$x^2_2 = \frac{208 - 208.01}{2} = -0.005,$$ negative discarded. 22. **Use $x^2 = 208.005$:** $$x = \sqrt{208.005} \approx 14.42.$$ 23. **Find $y$:** $$y = x - \frac{64}{x} = 14.42 - \frac{64}{14.42} = 14.42 - 4.44 = 9.98.$$ 24. **Answer:** Length $BC$ is approximately $9.98$ rounded to nearest hundredth. **Final answer:** $$\boxed{9.98}.$$