Giải thích các bước giải:

1.Xét $\Delta ABD,\Delta AHD$ có:

Chung $\hat D$

$\widehat{BAD}=\widehat{AHD}(=90^o)$

$\to \Delta ABD\sim\Delta HAD(g.g)$

2.Từ 1 $\to \dfrac{AD}{HD}=\dfrac{BD}{AD}$

$\to DA^2=DH.DB$

4.Ta có:

$AD=BC=9$

$DB=\sqrt{AB^2+AD^2}=15$

$DH=\dfrac{AD^2}{DB}=\dfrac{9^2}{15}=5.4$

$AH=\sqrt{AD^2-DH^2}=\sqrt{9^2-5.4^2}=7.2$

5.Ta có: $\Delta BHA\sim\Delta BAD(g.g)$

$\to \dfrac{S_{BHA}}{S_{BAD}}=\dfrac{AB^2}{BD^2}=\dfrac{12^2}{15^2}=\dfrac{16}{25}$

Mà $S_{BAD}=S_{BCD}$

$\to \dfrac{S_{BHA}}{S_{BCD}}=\dfrac{16}{25}$