áp dụng đinh lí pi-ta-go, ta tính được BC=20cm (1)

mà \(\dfrac{AB}{AC}=\dfrac{BD}{CD}\)( phân giác AD)\(\Leftrightarrow\dfrac{BD}{CD}=\dfrac{12}{16}=\dfrac{3}{4}\) (2)

từ (1),(2)\(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{60}{7}\\CD=\dfrac{80}{7}\end{matrix}\right.\)(3)

ta có \(AD=\dfrac{AB.AC}{BD}=9,6\)(4)

từ (3),(4)\(\Rightarrow\left\{{}\begin{matrix}S_{ABD}=\dfrac{288}{7}\\S_{ACD}=\dfrac{384}{7}\end{matrix}\right.\)\(\Rightarrow\dfrac{S_{ABD}}{S_{ACD}}=\dfrac{3}{4}\)

ta có: \(x+2y=1\Rightarrow x=1-2y\)

thay vào A, ta có:

\(A=\left(1-2y\right)^2+2y^2\)

\(A=\left(6y^2-4y+\dfrac{2}{3}\right)+\dfrac{1}{3}\)

\(A=\left(\sqrt{6}y-\dfrac{\sqrt{6}}{3}\right)^2+\dfrac{1}{3}\)

vậy A_min=1/3