1. **Problem statement:** Given an isosceles triangle $\triangle ABC$ with $|AB|=|AC|=6$ cm and $|BC|=4$ cm, find the points equidistant from $B$ and $C$ and 3 cm from $A$, then measure their distance from line $BC$. 2. **Key concepts and formulas:** - The locus of points equidistant from $B$ and $C$ is the perpendicular bisector of segment $BC$. - The locus of points 3 cm from $A$ is a circle centered at $A$ with radius 3 cm. - The intersection of these two loci gives the points satisfying both conditions. 3. **Step-by-step solution:** 1. Calculate the midpoint $M$ of $BC$: $$M = \left(\frac{B_x + C_x}{2}, \frac{B_y + C_y}{2}\right)$$ Since $|BC|=4$ cm, place $B$ at $(0,0)$ and $C$ at $(4,0)$ for convenience, so $$M = (2,0)$$ 2. The perpendicular bisector of $BC$ is the vertical line through $M$: $$x=2$$ 3. The circle centered at $A$ with radius 3 cm: Since $|AB|=6$ and $|AC|=6$, $A$ lies on the perpendicular bisector of $BC$ at a height $h$ such that $$AB^2 = AM^2 + BM^2 \Rightarrow 6^2 = h^2 + 2^2 \Rightarrow 36 = h^2 + 4 \Rightarrow h^2 = 32 \Rightarrow h = 4\sqrt{2}$$ So $A = (2, 4\sqrt{2})$. 4. Equation of the circle: $$ (x-2)^2 + (y - 4\sqrt{2})^2 = 3^2 = 9 $$ 5. Find intersection points of the circle and line $x=2$: Substitute $x=2$ into the circle equation: $$ (2-2)^2 + (y - 4\sqrt{2})^2 = 9 \Rightarrow (y - 4\sqrt{2})^2 = 9 $$ 6. Solve for $y$: $$ y - 4\sqrt{2} = \pm 3 \Rightarrow y = 4\sqrt{2} \pm 3 $$ 7. The two points are: $$ P_1 = (2, 4\sqrt{2} + 3), \quad P_2 = (2, 4\sqrt{2} - 3) $$ 8. Distance from $BC$ (the $x$-axis) is the absolute $y$-coordinate: $$ d_1 = 4\sqrt{2} + 3, \quad d_2 = 4\sqrt{2} - 3 $$ **Final answer:** The points equidistant from $B$ and $C$ and 3 cm from $A$ lie on $x=2$ at heights $4\sqrt{2} \pm 3$ cm, and their distances from $BC$ are $4\sqrt{2} + 3$ cm and $4\sqrt{2} - 3$ cm respectively.