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

* Hướng dẫn giải

Ta có: $\frac{AD}{AB}=\frac{DC}{BC}$ (t/c)

$\Rightarrow \frac{AD}{4}=\frac{DC}{6}=\frac{AD+DC}{4+6}=\frac{5}{10}=\frac{1}{2}$

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

Suy ra:

$$\begin{gathered} \frac{DI}{IB}=\frac{DC}{CB}=\frac{3}{6}=\frac{1}{2}\Rightarrow \frac{DI}{DB}=\frac{1}{3} \\ \frac{BE}{EA}=\frac{BC}{AC}=\frac{6}{5}\Rightarrow \frac{BE}{BA}=\frac{6}{11} \\ \frac{AD}{DC}=\frac{2}{3}\Rightarrow \frac{AD}{AC}=\frac{2}{5} \end{gathered}$$

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

Đáp án: A