Đáp án:

a) Theo Pytago ta có:

$\begin{array}{l}
B{C^2} = A{B^2} + A{C^2} = {12^2} + {16^2} = 400\\
 \Rightarrow BC = 20\left( {cm} \right)
\end{array}$

b) Xét ΔABC và ΔHBA có:

+ góc B chung

+ góc BAC = góc BHA = 90 độ

=> ΔABC ~ ΔHBA (g-g)

=> AB/BH = BC/AB

=> AB^2 = BH.BC

c)

Ta có:

$\begin{array}{l}
{S_{ABC}} = \frac{1}{2}.AB.AC = \frac{1}{2}.AH.BC\\
 \Rightarrow AH = \frac{{AB.AC}}{{BC}} = \frac{{12.16}}{{20}} = 9,6\left( {cm} \right)\\
Theo\,Pytago:A{B^2} = B{H^2} + A{H^2}\\
 \Rightarrow B{H^2} = {12^2} - 9,{6^2} = 51,84\\
 \Rightarrow BH = 7,2\left( {cm} \right)
\end{array}$

d) Áp dụng t/c đường phân giác ta có:

$\begin{array}{l}
\frac{{BD}}{{CD}} = \frac{{AB}}{{AC}} = \frac{{12}}{{16}} = \frac{3}{4}\\
 \Rightarrow \frac{{{S_{ABD}}}}{{{S_{ACD}}}} = \frac{{\frac{1}{2}.AH.BD}}{{\frac{1}{2}.AH.CD}} = \frac{{BD}}{{CD}} = \frac{3}{4}
\end{array}$