Gọi đường cao tứ giác $BDEC$ là $KH$
$→3AH=KH$ hay $\dfrac{AH}{KH}=3$
$→KH=\dfrac{1}{3}AH$
$DE//BC→DK//BH$
$→\dfrac{AB}{BD}=\dfrac{AH}{KH}=3$ (Định lý Talet)
$DE//BC→EK//CH$
$→\dfrac{AC}{AE}=\dfrac{AH}{KH}=3$ (Định lý Talet)
$DE//BC$
$→\dfrac{AB}{AD}=\dfrac{AC}{AE}=\dfrac{BC}{DE}=3$ (Định lý Talet)
$→DE=\dfrac{1}{3}BC$
$S_{ΔABC}=\dfrac{1}{2}.AH.BC$
$S_{BDEC}=\dfrac{1}{2}(DE+BC).KH\\=\dfrac{1}{2}(\dfrac{1}{3}BC+BC).\dfrac{1}{3}AH\\=\dfrac{1}{6}.\dfrac{4}{3}BC.AH\\=\dfrac{2}{9}.BC.AH=200\\↔BC.AH=900(cm)\\→S_{ΔABC}=\dfrac{1}{2}.BC.AH=\dfrac{1}{2}.900=450(cm^2)$
Vậy $S_{ΔABC}=450cm^2$
