cho tam giác ABC. M,N thuộc 2 cạnh AB,AC. CM: (SAMN)/(SABC) = (AM)/(AB) . (AN)/(AC)

Kẻ $BH\bot AC$ và $MK\bot AC$

$\to BH//MK$

$\to \dfrac{MK}{BH}=\dfrac{AM}{AB}$ (hệ quả định lý Ta-let)

Ta có:

${{S}_{\Delta AMN}}=\dfrac{1}{2}MK\cdot AN$

${{S}_{\Delta ABC}}=\dfrac{1}{2}BH\cdot AC$

$\to \dfrac{{{S}_{\Delta AMN}}}{{{S}_{\Delta ABC}}}=\dfrac{MK}{BH}\cdot \dfrac{AN}{AC}$

$\to \dfrac{{{S}_{\Delta AMN}}}{{{S}_{\Delta ABC}}}=\dfrac{AM}{AB}\cdot \dfrac{AN}{AC}$