mn ơi giúp mk bài số 7 vs ak
View attachment 197459
Kẻ $MK \perp BC, AH \perp BC$, ta có:
$S_{BMC}=\dfrac12 MK.BC$
$S_{ABC}=\dfrac12 AH.BC$
$\implies \dfrac{S_{BMC}}{S_{ABC}}=\dfrac{MK}{AH}$
Vì $MK//AH \implies \dfrac{MK}{AH}=\dfrac{MD}{AD}$
Do đó: $\dfrac{S_{BMC}}{S_{ABC}}=\dfrac{MD}{AD}$
Chứng minh tương tự ta có:
$\dfrac{S_{AMC}}{S_{ABC}}=\dfrac{ME}{BE}$
$\dfrac{S_{AMB}}{S_{ABC}}=\dfrac{MF}{CF}$
Vậy $\dfrac{MD}{AD}+\dfrac{ME}{BE}+\dfrac{MF}{CF}=\dfrac{S_{BMC}}{S_{ABC}}+\dfrac{S_{AMC}}{S_{ABC}}+\dfrac{S_{AMB}}{S_{ABC}}=1$
Ôn thi học kì