Đáp án:
a)
$\begin{array}{l}
A{B^2} = A{H^2} + B{H^2} = {12^2} + {9^2} = 225\\
HC = BC - BH = 16\left( {cm} \right)\\
\Rightarrow A{C^2} = H{C^2} + A{H^2} = {16^2} + {12^2} = 400\\
\Rightarrow A{B^2} + A{C^2} = 225 + 400 = 625 = B{C^2}\\
\Rightarrow A{B^2} + A{C^2} = B{C^2}
\end{array}$
Vậy tam giác ABC vuông tại A
b)
Áp dụng Talet trong tam giác HAC có: BD//AC
$\begin{array}{l}
\Rightarrow \dfrac{{HD}}{{AH}} = \dfrac{{BH}}{{HC}} = \dfrac{{BD}}{{AC}}\\
\Rightarrow \dfrac{{HD}}{{12}} = \dfrac{9}{{16}} = \dfrac{{BD}}{{20}}\\
\Rightarrow \left\{ \begin{array}{l}
HD = 6,75\left( {cm} \right)\\
BD = 11,25\left( {cm} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
B{H^2} = 81\\
AC.BD = 20.11,25 = 81
\end{array} \right.\\
\Rightarrow B{H^2} = AC.BD\\
c)DE \bot AC\\
\Rightarrow DE//AB\\
\Rightarrow \dfrac{{HD}}{{AH}} = \dfrac{{HF}}{{BH}}\\
\Rightarrow \dfrac{{6,75}}{{12}} = \dfrac{{HF}}{9}\\
\Rightarrow HF = 5,625\left( {cm} \right)\\
\Rightarrow \left\{ \begin{array}{l}
HF.HC = 5,625.16 = 81\\
B{H^2} = 81
\end{array} \right.\\
\Rightarrow B{H^2} = HF.HC
\end{array}$
d) Xét ΔAHC và ΔDHB có:
+ góc AHC = góc DHB = 90 độ
+ góc ACH = góc DBH (so le trong)
=> ΔAHC ~ ΔDHB (g-g)
$\begin{array}{l}
\Rightarrow \dfrac{{AH}}{{DH}} = \dfrac{{CH}}{{HB}}\\
\Rightarrow AH.HB = DH.CH\\
\Rightarrow \dfrac{1}{2}.AH.HB = \dfrac{1}{2}.DH.CH\\
\Rightarrow {S_{AHB}} = {S_{DCH}}
\end{array}$
