Cho tam giác ABC có: AB = 4cm, AC = 5cm, BC = 6cm. Các đường phân giác BD

* Hướng dẫn giải

Đáp án A

Ta có: $\frac{AD}{AB}=\frac{DC}{BC}$ (t/c) $\Rightarrow \frac{AD}{4}=\frac{DC}{6}=\frac{AD+DC}{4+6}=\frac{5}{10}=\frac{1}{2}$

$\Rightarrow AD = 4.\frac{1}{2}= 2, DC = 6.\frac{1}{2} = 3$

Suy ra $\frac{DI}{IB}=\frac{DC}{CB}=\frac{3}{6}=\frac{1}{2}\Rightarrow \frac{DI}{DB}=\frac{1}{3}$

$\frac{BE}{EA}=\frac{BC}{AC}=\frac{6}{5}\Rightarrow \frac{BE}{BA}=\frac{6}{11}$

$\frac{AD}{DC}=\frac{2}{3}\Rightarrow \frac{AD}{AC}=\frac{2}{5}$

Suy ra $S_{DIE} = \frac{1}{3}{S}_{BDE}$

$\Rightarrow S_{DIE} =\frac{1}{3}.\frac{6}{11}.\frac{2}{5}=\frac{4}{55} S_{ABC}$

Vậy $\frac{S_{DIE}}{S_{ABC}}=\frac{4}{55}$