1. **Problem Statement:** We have a semicircle centered at point $O$ with radius 8 cm. Points $A$ and $B$ are endpoints of the diameter. Point $C$ lies on the semicircle. We need to find the location of point $D$ such that: - $D$ is equidistant from lines $BA$ and $BC$. - $D$ is 8 cm away from $O$. 2. **Understanding the problem:** - Since $D$ is equidistant from lines $BA$ and $BC$, it lies on the angle bisector of angle $ABC$. - Since $D$ is 8 cm from $O$, it lies on the circle centered at $O$ with radius 8 cm. 3. **Formulas and rules:** - The locus of points equidistant from two lines is the angle bisector of the angle formed by those lines. - The locus of points at a fixed distance $r$ from a point $O$ is a circle with radius $r$ centered at $O$. 4. **Step-by-step solution:** - Identify the angle bisector of angle $ABC$. - The point $D$ lies on this bisector. - Since $D$ is also 8 cm from $O$, $D$ lies on the circle centered at $O$ with radius 8 cm. - Therefore, $D$ is the intersection of the angle bisector of $\angle ABC$ and the circle with radius 8 cm centered at $O$. 5. **Summary:** - Draw the angle bisector of $\angle ABC$. - Draw the circle centered at $O$ with radius 8 cm. - The intersection point(s) of these two loci is the location of $D$. **Final answer:** Point $D$ is located at the intersection of the angle bisector of $\angle ABC$ and the circle centered at $O$ with radius 8 cm. This uses the concept of loci: the angle bisector is the locus of points equidistant from $BA$ and $BC$, and the circle is the locus of points 8 cm from $O$.