If the angle bisector $AD$ of the angle $A$ of the triangle $ABC$ divides the side $BC$ into two segments $BD = 4, DC = 2$, then:
Step-by-Step Solution
Key Concept: The angle bisector theorem establishes the relationship $c = 2b$, and triangle inequalities on this constraint determine the valid ranges for $b$ and $c$.
By the angle bisector theorem, $\frac{AB}{AC} = \frac{BD}{DC} = \frac{4}{2} = 2$, so $c = 2b$. Since $BC = 6$ is fixed, the triangle inequality gives: $b + c > 6 \Rightarrow b + 2b > 6 \Rightarrow b > 2$; $b + 6 > c \Rightarrow b + 6 > 2b \Rightarrow b < 6$; and $c + 6 > b$ is always satisfied. Thus $2 < b < 6$ and $4 < c < 12$. The altitude from $A$ is $h = \frac{2\cdot\text{Area}}{BC} = \frac{2\cdot\text{Area}}{6}$. Using Heron's formula with $s = \frac{b+c+6}{2} = \frac{3b+6}{2}$, the area is maximized when the triangle is right-angled at $A$ (or approaches it), giving maximum altitude $h_{\max} = 4$ when $b = c = 2\sqrt{5}$ (approximately 4.47), but checking boundary behavior as $b \to 2^+$ or $b \to 6^-$, the altitude reaches maximum value of 4.
Correct Answer: 1,2,4