Cách 1:  Đường trung bình – Ta-let

Kẻ $MH//d\,\,\left( H\in AC \right)$

Kẻ $BG//d\,\,\,\left( G\in AC \right)$

$\Rightarrow BG//MH$

Xét $\Delta GBC$, ta có:

+  $M$ trung điểm $BC$

+  $MH//BG$

Nên $H$ là trung điểm $CG$

Do đó $2GH=GC$

Xét $\Delta AMH$ có $FE//MH$:

$\Rightarrow \dfrac{AM}{AF}=\dfrac{AH}{AE}$ (Định lý Ta-let)

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{2AH}{AE}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{2\left( AG+GH \right)}{AE}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{2AG+2GH}{AE}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AG+AG+GC}{AE}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AG+AC}{AE}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AG}{AE}+\dfrac{AC}{AE}$

Mà: $\dfrac{AG}{AE}=\dfrac{AB}{AD}$ (Vì $DE//BG$, định lý Ta-let)

Vậy: $\dfrac{2AM}{AF}=\dfrac{AB}{AD}+\dfrac{AC}{AE}$

Cách 2:  Tam giác bằng nhau – Ta-let

Kẻ $BG//d\,\,\left( G\in AM \right)$

Kẻ $CH//d\,\,\left( H\in AM \right)$

$\Rightarrow BG//CH$

Xét $\Delta GBM$ và $\Delta HCM$, ta có:

+  $\widehat{GBM}=\widehat{HCM}$ (vì $BG//CH$, hai góc so le trong)

+  $BM=CM$ (vì $M$ là trung điểm $BC$)

+  $\widehat{GMB}=\widehat{HMC}$ (hai góc đối đỉnh)

Nên $\Delta GBM=\Delta HCM\left( g.c.g \right)$

Do đó $GM=HM$ (hai cạnh tương ứng)

Vậy $HM-GM=0$

Áp dụng định lý Ta-let, ta có:

+  $\Delta ABG$ với $DF//BG\Rightarrow \dfrac{AB}{AD}=\dfrac{AG}{AF}$

+  $\Delta ACH$ với $EF//CH\Rightarrow \dfrac{AC}{AE}=\dfrac{AH}{AF}$

$\Rightarrow \dfrac{AB}{AD}+\dfrac{AC}{AE}=\dfrac{AG}{AF}+\dfrac{AH}{AF}$

$\Rightarrow \dfrac{AB}{AD}+\dfrac{AC}{AE}=\dfrac{AM-GM}{AF}+\dfrac{AM+HM}{AF}$

$\Rightarrow \dfrac{AB}{AD}+\dfrac{AC}{AE}=\dfrac{AM}{AF}+\dfrac{AM}{AF}+\dfrac{HM-GM}{AF}$

$\Rightarrow \dfrac{AB}{AD}+\dfrac{AC}{AE}=\dfrac{2AM}{AF}+\dfrac{0}{AF}$

$\Rightarrow \dfrac{AB}{AD}+\dfrac{AC}{AE}=\dfrac{2AM}{AF}$

Cách 3:  Công thức diện tích $S=\dfrac{1}{2}ab\sin \alpha $

Vì $M$ là trung điểm $BC$

Nên ${{S}_{\Delta ABM}}={{S}_{\Delta ACM}}=\dfrac{1}{2}{{S}_{\Delta ABC}}$

Ta có:

$\begin{cases}{{S}_{\Delta ADE}}=\dfrac{1}{2}AD.AE.\sin \widehat{BAC}\\{{S}_{\Delta ABC}}=\dfrac{1}{2}AB.AC.\sin \widehat{BAC}\end{cases}\Rightarrow \dfrac{{{S}_{\Delta ADE}}}{{{S}_{\Delta ABC}}}=\dfrac{AD.AE}{AB.AC}\,\,\,\,\,\left( 1 \right)$

+  $\begin{cases}{{S}_{\Delta ADF}}=\dfrac{1}{2}AD.AF.\sin \widehat{{{A}_{1}}}\\{{S}_{\Delta ABM}}=\dfrac{1}{2}AB.AM.\sin \widehat{{{A}_{1}}}\end{cases}\Rightarrow\dfrac{{{S}_{\Delta ADF}}}{{{S}_{\Delta ABM}}}=\dfrac{AD.AF}{AB.AM}$

+ $\begin{cases}{{S}_{\Delta AEF}}=\dfrac{1}{2}AE.AF\sin \widehat{{{A}_{2}}}\\{S}_{\Delta ACM}=\dfrac{1}{2}AC.AM.\sin \widehat{{{A}_{2}}}\end{cases}\Rightarrow\dfrac{{{S}_{\Delta AEF}}}{{{S}_{\Delta ACM}}}=\dfrac{AE.AF}{AC.AM}$

$\Rightarrow \dfrac{{{S}_{\Delta ADF}}}{{{S}_{\Delta ABM}}}+\dfrac{{{S}_{\Delta AEF}}}{{{S}_{\Delta ACM}}}=\dfrac{AD.AF}{AB.AM}+\dfrac{AE.AF}{AC.AM}$

$\Rightarrow \dfrac{{{S}_{\Delta ADF}}+{{S}_{\Delta AEF}}}{{{S}_{\Delta ABM}}}=\dfrac{AF}{AM}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)$   (Vì ${{S}_{\Delta ABM}}={{S}_{\Delta ACM}}$)

$\Rightarrow \dfrac{{{S}_{\Delta ADE}}}{{{S}_{\Delta ABM}}}=\dfrac{AF}{AM}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)$

$\Rightarrow \dfrac{{{S}_{\Delta ADE}}}{\dfrac{1}{2}{{S}_{\Delta ABC}}}=\dfrac{AF}{AM}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)$

$\Rightarrow \dfrac{{{S}_{\Delta ADE}}}{{{S}_{\Delta ABC}}}=\dfrac{AF}{2AM}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)\,\,\,\,\,\left( 2 \right)$

Từ $\left( 1 \right)$ và $\left( 2 \right)$

Ta được $\dfrac{AD.AE}{AB.AC}=\dfrac{AF}{2AM}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AB.AC}{AD.AE}\left( \dfrac{AD}{AB}+\dfrac{AE}{AC} \right)$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AB.AC}{AD.AE}\cdot \dfrac{AD}{AB}\,\,+\,\,\dfrac{AB.AC}{AD.AE}\cdot \dfrac{AE}{AC}$

$\Rightarrow \dfrac{2AM}{AF}=\dfrac{AC}{AE}+\dfrac{AB}{AD}$