To solve the given problem, we can use the formula for the area of a triangle inscribed in a circle with given properties:

Problem Breakdown:

1. Given:

   • Radius of incircle $r = 4 \, \text{cm}$
   • Area of triangle $\Delta = 84 \, \text{cm}^2$
   • Two sides divided by the incenter:
     • $D$ divides $BC$ into $BD = 8 \, \text{cm}$ and $DC = 6 \, \text{cm}$.

2. To Find:

   • Sides $AB$ and $AC$.

3. Key Formulae:

   • Area $\Delta = r \cdot s$, where $s$ is the semi-perimeter.
   • Semi-perimeter $s = \frac{AB + BC + AC}{2}$.

Steps:

1. Calculate the semi-perimeter $s$: Using $\Delta = r \cdot s$: $84 = 4 \cdot s \implies s = \frac{84}{4} = 21 \, \text{cm}.$

2. Relation for the sides: Let $AB = x$ and $AC = y$, and $BC = BD + DC = 8 + 6 = 14 \, \text{cm}$.

   Using the semi-perimeter formula: $s = \frac{x + y + 14}{2}.$ Substituting $s = 21$: $21 = \frac{x + y + 14}{2} \implies x + y + 14 = 42 \implies x + y = 28.$

3. Split the sides using the incenter property: Using the property of tangents:

   • Tangents from a point to a circle are equal. So:
     • $AB = x = BD + BA = 8 + z$,
     • $AC = y = DC + CA = 6 + z$.

   Therefore: $x = 8 + z \quad \text{and} \quad y = 6 + z.$

4. Solve for $z$: From $x + y = 28$: $(8 + z) + (6 + z) = 28 \implies 14 + 2z = 28 \implies 2z = 14 \implies z = 7.$

5. Find $x$ and $y$: Substituting $z = 7$: $x = 8 + 7 = 15, \quad y = 6 + 7 = 13.$

Final Answer:

The lengths of the sides are: $AB = 15 \, \text{cm}, \quad AC = 13 \, \text{cm}, \quad BC = 14 \, \text{cm}.$

Would you like further explanation on any step?

Here are related questions you might explore:

1. How does the incenter property simplify triangle geometry problems?
2. What is the significance of the semi-perimeter in such calculations?
3. How are tangent properties applied to triangle and circle problems?
4. Can this problem be solved graphically?
5. What are some other applications of incircles in triangle geometry?

Tip: Always start such problems by identifying key relationships like tangents and semi-perimeters.