Qua N kẻ đường thẳng song song CM cắt AB tại D
\(\Rightarrow\widehat{BND}=\widehat{ACM}\) (góc có cạnh tương ứng song song)
\(\Rightarrow\Delta_VBND\sim\Delta_VACM\left(g.g\right)\)
\(\Rightarrow\dfrac{BD}{BN}=\dfrac{AM}{AC}=\dfrac{1}{3}\Rightarrow BD=\dfrac{1}{3}BN=\dfrac{1}{6}AB\)
\(\Rightarrow MD=AB-AM-BD=AB-\dfrac{1}{3}AB-\dfrac{1}{6}AB=\dfrac{1}{2}AB\)
Áp dụng Talet: \(\dfrac{AF}{BN}=\dfrac{AE}{NE}=\dfrac{AM}{MD}=\dfrac{\dfrac{1}{3}AB}{\dfrac{1}{2}AB}=\dfrac{2}{3}\)
\(\Rightarrow AF=\dfrac{2}{3}BN=\dfrac{2}{3}.\dfrac{1}{2}AB=\dfrac{1}{3}AB=AM\left(đpcm\right)\)
b.
Ta có: \(CF=AF+AC=\dfrac{1}{3}AB+AB=\dfrac{4}{3}AB\)
\(BF=\sqrt{AB^2+AF^2}=\sqrt{AB^2+\left(\dfrac{1}{3}AB\right)^2}=\dfrac{AB\sqrt{10}}{3}\)
\(\dfrac{EM}{ND}=\dfrac{AM}{AD}=\dfrac{\dfrac{1}{3}AB}{\dfrac{5}{6}AB}=\dfrac{2}{5}\Rightarrow EM=\dfrac{2}{5}ND=\dfrac{AB\sqrt{10}}{15}\)
\(\dfrac{BE}{EF}=\dfrac{BN}{AF}=\dfrac{\dfrac{1}{2}AB}{\dfrac{1}{3}AB}=\dfrac{3}{2}\Rightarrow\dfrac{BE}{BF-BE}=\dfrac{3}{2}\)
\(\Rightarrow BE=\dfrac{3}{5}BF=\dfrac{AB\sqrt{10}}{5}\)
\(\Rightarrow BE^2+EM^2=\dfrac{2}{5}AB^2+\dfrac{2}{45}AB^2=\dfrac{4}{9}AB^2=\left(\dfrac{2}{3}AB\right)^2=BM^2\)
\(\Rightarrow\Delta BEM\) vuông tại E \(\Rightarrow\Delta CEF\) vuông tại E
\(\Rightarrow EH\) là trung tuyến ứng với cạnh huyền
\(\Rightarrow EH=\dfrac{1}{2}CF=\dfrac{2}{3}AB=BM\)
