Giải thích các bước giải:
a.Ta có: $$\Delta HAB, \Delta AHC$$ vuông tại $$H, HM\perp AB, HN\perp AC$$
$$\to AM\cdot AB=AH^2=AN\cdot AC$$
b.Xét $$\Delta AMN,\Delta ABC$$ có:
Chung $$\hat A$$
$$\dfrac{AM}{AC}=\dfrac{AN}{AB}$$ vì $$AM\cdot AB=AN\cdot AC$$
$$\to\Delta ANM\sim\Delta ABC(c.g.c)$$
$$\to\widehat{AMN}=\widehat{ACB}$$
c.Từ câu c
$$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{MN}{BC})^2$$
Vì $$HM\perp AB , HN\perp AC, AB\perp AC\to AMHN$$ là hình chữ nhật
$$\to AH=MN$$
$$\to \dfrac{S_{AMN}}{S_{ABC}}=(\dfrac{AH}{BC})^2$$
Ta có:
$$\sin^2B\cdot \sin^2C$$
$$=(\dfrac{AC}{BC})^2\cdot(\dfrac{AB}{BC})^2$$
$$=\dfrac{(AC\cdot AB)^2}{BC^4}$$
$$=\dfrac{(AH\cdot BC)^2}{BC^4}$$
$$=\dfrac{AH^2}{BC^2}$$
$$=\dfrac{MN^2}{BC^2}$$
$$\to \dfrac{S_{AMN}}{S_{ABC}}=\sin^2B\cdot \sin^2C$$