a) -Xét △ABM và △ACN có:
\(\widehat{AMB}=\widehat{ANC}=90^0\)
\(\widehat{BAC}\) là góc chung.
\(\Rightarrow\)△ABM∼△ACN (g-g).
\(\Rightarrow\dfrac{AB}{AC}=\dfrac{AM}{AN}\Rightarrow\dfrac{AB}{AM}=\dfrac{AC}{AN}\) nên \(AN.AB=AM.AC\)
b) -Xét △AMN và △ABC có:
\(\widehat{BAC}\) là góc chung
\(\dfrac{AB}{AM}=\dfrac{AC}{AN}\left(cmt\right)\)
\(\Rightarrow\)△AMN∼△ABC (c-g-c).
c) -Có: △AMB vuông tại M và \(\widehat{MAB}=60^0\left(gt\right)\)
\(\Rightarrow AM=\dfrac{1}{2}AB\)
-Có: △AMN∼△ABC (cmt)
\(\Rightarrow\dfrac{S_{ABC}}{S_{AMN}}=\left(\dfrac{AB}{AM}\right)^2=\left(\dfrac{2AM}{AM}\right)^2=4\)
\(\Rightarrow S_{ABC}=4S_{AMN}\)
\(\Rightarrow\)
