1. **Problem statement:** We have a right-angled triangle BAC with right angle at A. Given lengths are $AB=12$ and $CD=1$. A semicircle with diameter $AD$ is tangent to side $BC$. We need to find the radius of this semicircle. 2. **Set up the coordinate system:** Place point $A$ at the origin $(0,0)$. 3. Since $AB=12$ and $AB$ is horizontal, place $B$ at $(12,0)$. 4. Since $AC$ is vertical (right angle at $A$), let $C$ be at $(0,h)$ for some $h$. 5. Point $D$ lies on $AC$ such that $CD=1$, so $D$ is at $(0,h-1)$. 6. The semicircle has diameter $AD$, so its center $O$ is midpoint of $A(0,0)$ and $D(0,h-1)$: $$O = \left(0, \frac{h-1}{2}\right)$$ 7. The radius $r$ of the semicircle is half the length of $AD$: $$r = \frac{|AD|}{2} = \frac{h-1}{2}$$ 8. The semicircle is tangent to $BC$. The line $BC$ passes through $B(12,0)$ and $C(0,h)$. 9. Equation of line $BC$: $$y = m x + c$$ Slope: $$m = \frac{h-0}{0-12} = -\frac{h}{12}$$ Intercept $c$ is $h$ since at $x=0$, $y=h$. So, $$y = -\frac{h}{12} x + h$$ 10. The distance from center $O(0,\frac{h-1}{2})$ to line $BC$ must equal the radius $r$ for tangency: Distance formula from point $(x_0,y_0)$ to line $Ax+By+C=0$ is: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ Rewrite line $BC$ in standard form: $$\frac{h}{12} x + y - h = 0$$ So, $$A = \frac{h}{12}, B=1, C=-h$$ Distance $d$ from $O$ to $BC$: $$d = \frac{\left| \frac{h}{12} \cdot 0 + 1 \cdot \frac{h-1}{2} - h \right|}{\sqrt{\left(\frac{h}{12}\right)^2 + 1^2}} = \frac{\left| \frac{h-1}{2} - h \right|}{\sqrt{\frac{h^2}{144} + 1}} = \frac{\left| \frac{h-1-2h}{2} \right|}{\sqrt{\frac{h^2}{144} + 1}} = \frac{\left| \frac{-h-1}{2} \right|}{\sqrt{\frac{h^2}{144} + 1}} = \frac{\frac{h+1}{2}}{\sqrt{\frac{h^2}{144} + 1}}$$ 11. Set distance equal to radius: $$\frac{h+1}{2 \sqrt{\frac{h^2}{144} + 1}} = \frac{h-1}{2}$$ Multiply both sides by $2$: $$\frac{h+1}{\sqrt{\frac{h^2}{144} + 1}} = h-1$$ 12. Square both sides: $$\frac{(h+1)^2}{\frac{h^2}{144} + 1} = (h-1)^2$$ Multiply both sides by denominator: $$(h+1)^2 = (h-1)^2 \left( \frac{h^2}{144} + 1 \right)$$ 13. Expand: $$(h+1)^2 = (h-1)^2 \left( \frac{h^2}{144} + 1 \right)$$ $$(h+1)^2 = (h-1)^2 \frac{h^2}{144} + (h-1)^2$$ 14. Rearrange: $$(h+1)^2 - (h-1)^2 = (h-1)^2 \frac{h^2}{144}$$ 15. Use difference of squares: $$(h+1)^2 - (h-1)^2 = [(h+1)-(h-1)] \cdot [(h+1)+(h-1)] = (2)(2h) = 4h$$ So: $$4h = (h-1)^2 \frac{h^2}{144}$$ Multiply both sides by 144: $$576 h = (h-1)^2 h^2$$ Divide both sides by $h$ (assuming $h \neq 0$): $$576 = (h-1)^2 h$$ 16. Let $x = h$, then: $$576 = (x-1)^2 x = x (x-1)^2 = x (x^2 - 2x + 1) = x^3 - 2x^2 + x$$ Rewrite: $$x^3 - 2x^2 + x - 576 = 0$$ 17. Try integer roots using Rational Root Theorem. Test $x=8$: $$8^3 - 2 \cdot 8^2 + 8 - 576 = 512 - 128 + 8 - 576 = (512 - 128) + (8 - 576) = 384 - 568 = -184 \neq 0$$ Try $x=9$: $$729 - 162 + 9 - 576 = 729 - 162 + 9 - 576 = (729 - 162) + (9 - 576) = 567 - 567 = 0$$ So $x=9$ is a root. 18. Since $h=9$, the radius is: $$r = \frac{h-1}{2} = \frac{9-1}{2} = \frac{8}{2} = 4$$ **Final answer:** The radius of the semicircle is $\boxed{4}$.