Xét $ΔABD$ và $ΔABC$ có:

Chung chiều cao hạ từ $A$

$⇒\dfrac{S_{ABD}}{S_{ABC}}=\dfrac{BD}{BC}$

$AD$ phân giác $\widehat{BAC}$

$⇒\dfrac{CD}{BD}=\dfrac{AC}{AB}$

$⇒\dfrac{CD+BD}{BD}=\dfrac{AC+AB}{AB}$

$⇒\dfrac{BC}{BD}=\dfrac{AC+AB}{AB}$

$⇒\dfrac{BD}{BC}=\dfrac{AB}{AC+AB}$

$⇒\dfrac{S_{ABD}}{S_{ABC}}=\dfrac{BD}{BC}=\dfrac{AB}{AC+AB}$

$⇒S_{ABD}=\dfrac{4}{4+7}.S_{ABC}$

$⇒S_{ABD}=\dfrac{4}{11}.S_{ABC}$

Ta có: 

$\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{BM}{BC}=\dfrac{1}{2}$ ( $M$ là trung điểm $BC$)

$⇒S_{ABM}=\dfrac{1}{2}.S_{ABC}$

$⇒S_{ADM}=S_{ABM}-S_{ABD}=\dfrac{1}{2}.S_{ABC}-\dfrac{4}{11}.S_{ABC}=\dfrac{3}{22}.S_{ABC}$

Kẻ `AH\bot BC(H\in BC)`

`S_{AMC}=1/2 . AH . CM`

`S_{ABC}=1/2 . AH . BC`

`->S_{AMC}/S_{ABC}=(CM)/(BC)=1/2`

`->S_{AMC}=1/2 S_{ABC}`

`\triangle ABC` có `AD` là đường phân giác

`->(CD)/(BD)=(AC)/(AB)=7/4`

`-> (CD)/(BD+CD)=7/(4+7)=7/11`

`->(CD)/(BC)=7/11`

`S_{ABC}=1/2 . AH . BC`

`S_{ADC}=1/2 . AH . CD`

`->S_{ABC}/S_{ADC}=(BC)/(CD)=11/7`

`-> S_{ADC}=7/11 S_{ABC}`

`S_{AMC}+S_{ADM}=S_{ADC}`

`-> 1/2 S_{ABC}+S_{ADM}=7/11 S_{ABC}`

`->S_{ADM}=3/22 S_{ABC}`

`->S_{ADM}/S_{ABC}=3/22`