a) Ta có:
$+)\quad \sin\widehat{ABH}=\dfrac{AH}{AB}$
$\Rightarrow AH = AB.\sin\widehat{ABH}= 6.\sin45^\circ$
$\Rightarrow AH = 3\sqrt2$
$+)\quad \sin\widehat{ACH}=\dfrac{AH}{AC}$
$\Rightarrow AC =\dfrac{AH}{\sin\widehat{ACH}}=\dfrac{3\sqrt2}{\sin30^\circ}$
$\Rightarrow AC = 6\sqrt2$
Ta được:
$S_{ABC}=\dfrac12AB.AC.\sin\widehat{A}$
$\Rightarrow S_{ABC}=\dfrac12\cdot 6\cdot 6\sqrt2\cdot \sin105^\circ$
$\Rightarrow S_{ABC}= 9 + 9\sqrt3\ cm^2$
Hoặc:
$S_{ABC}=\dfrac12AH.BC$
$\Rightarrow S_{ABC}=\dfrac12AH.(BH + CH)$
$\Rightarrow S_{ABC}=\dfrac12AH(AB.\cos\widehat{B} + AC.\cos\widehat{C})$
$\Rightarrow S_{ABC}=\dfrac12\cdot 3\sqrt2\left(6.\cos45^\circ + 6\sqrt2.\cos30^\circ\right)$
$\Rightarrow S_{ABC}=9 + 9\sqrt3\ cm^2$
b) Ta có:
$ S_{AMN}=\dfrac12AM.AN.\sin\widehat{A}$
$\qquad =\dfrac{AM.AN}{AB.AC}\cdot \dfrac12AB.AC.\sin\widehat{A}$
$\qquad = \dfrac{AM.AN}{AB.AC}\cdot S_{ABC}$
$\qquad =\dfrac{AM}{AH}\cdot \dfrac{AN}{AH}\cdot \dfrac{AH}{AB}\cdot \dfrac{AH}{AC}\cdot S_{ABC}$
$\qquad = \sin\widehat{AHM}\cdot \sin\widehat{AHN}\cdot \sin\widehat{B}\cdot \sin\widehat{C}\cdot S_{ABC}$
$\qquad = \sin\widehat{B}\cdot \sin\widehat{C}\cdot \sin\widehat{B}\cdot \sin\widehat{C}\cdot S_{ABC}$
$\qquad = \sin^2\widehat{B}.\sin^2\widehat{C}.S_{ABC}$
