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In a triangle $$ABC$$, $$D$$ is a point on the side $$BC$$ such that $$AD$$ is the internal angle bisector of the angle $$A$$. The lengths of the sides $$AB$$, $$BC$$, $$AC$$ are 4, 6, 8 respectively. Then the area of triangle $$ACD$$ is
By Heron's formula with $$s = \frac{4 + 6 + 8}{2} = 9$$, area of $$\triangle ABC = \sqrt{9 \times 5 \times 3 \times 1} = 3\sqrt{15}$$. By the angle bisector theorem $$\frac{BD}{DC} = \frac{4}{8} = \frac{1}{2}$$, so $$DC = \frac{2}{3}BC$$. Triangles $$ACD$$ and $$ABC$$ have the same height from $$A$$, so area of $$\triangle ACD = \frac{2}{3} \times 3\sqrt{15} = 2\sqrt{15}$$.
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