$\tan{\widehat{BAH}}=\dfrac{HB}{HA} = \dfrac{1}{2} \Leftrightarrow \widehat{BAH} =\arctan{ \left ( \dfrac{1}{2} \right )} $
$\widehat{CAH} = 3 .\widehat{BAH} =3. \arctan{ \left ( \dfrac{1}{2} \right )}$
$S_{ABH}= \dfrac{1}{2} HA.HB = 9$
$\tan{\widehat{CAH}}=\dfrac{HC}{HA} \Leftrightarrow HC = HA. \tan{\widehat{CAH}} = 6 . \tan{\left ( 3. \arctan{ \left ( \dfrac{1}{2} \right )} \right )} = 33$
$S_{ACH} = \dfrac{1}{2} HA.HC= 99 $
Do đó $S_{ABC} = S_{ABH} + S_{ACH} = 108$