Bạn tự vẽ hình nha ^^
Kẻ DE//AB $\implies \dfrac{AE}{CE}=\dfrac{BD}{CD}$
AD là đường phân giác $\implies \dfrac{BD}{CD}=\dfrac{AB}{AC}$
$\implies \dfrac{AE}{CE}=\dfrac{AB}{AC}\implies \dfrac{AE}{CE+AE}=\dfrac{AB}{AC+AB}$
$\implies \dfrac{AE}{35}=\dfrac{14}{49}\implies AE=10$ (cm)
DE//AB $\implies \widehat{BAD}=\widehat{ADE}$ (so le trong)
mà $\widehat{BAD}=\widehat{CAD}\implies \widehat{ADE}=\widehat{CAD}$ hay $\widehat{ADE}=\widehat{EAD}\implies \triangle ADE$ cân tại E
Kẻ $EH\perp AD\implies AH=DH=6$
Theo ĐL Py-ta-go ta có: $EH^2=AE^2-AH^2=10^2-6^2=64\implies EH=8$ (cm)
$\implies S_{ADE}=\dfrac12.8.12=48(cm^2)$
Ta có:
$\dfrac{S_{ADE}}{S_{ADC}}=\dfrac{AE}{AC}=\dfrac{10}{35}=\dfrac{2}{7}\implies S_{ADC}=168(cm^2)$
$\dfrac{S_{ADC}}{S_{ADB}}=\dfrac{CD}{BD}=\dfrac{AC}{AB}=\dfrac{35}{14}=\dfrac{5}{2}\implies \dfrac{S_{ADC}}{S_{ADB}+S_{ADC}}=\dfrac{5}{2+5}=\dfrac{5}{7}\implies \dfrac{168}{S_{ABC}}=\dfrac{5}{7}\\\implies S_{ABC}=235,2(cm^2)$
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