Định lí Thalès trong tam giác
$\triangle ABC$ có $DE\parallel BC$ ($D\in AB$, $E\in AC$) thì
$\displaystyle \frac{AD}{AB}$ $\displaystyle =\frac{AE}{AC};\qquad\frac{AD}{DB}$ $\displaystyle =\frac{AE}{EC};\qquad\frac{DB}{AB}$ $\displaystyle =\frac{EC}{AC}$
- Đảo: nếu $\dfrac{AD}{DB}$ $=\dfrac{AE}{EC}$ thì $DE\parallel BC$.
- Hệ quả: $DE\parallel BC$ thì $\dfrac{AD}{AB}$ $=\dfrac{AE}{AC}$ $=\dfrac{DE}{BC}$.