Từ P kẻ PK vuông góc với CD ; PH vuông góc với AB

Đặt PK = m

      PH = n

Ta có m x DC = 2 x 2 = 4

           \(\Rightarrow m=\dfrac{4}{DC}\)

          n x AB = 2 x 4 = 8

        \(\Rightarrow n=\dfrac{8}{AB}\)

Lại có \(\left(m+n\right)\left(DC+AB\right)=2\left(2+3+4+5\right)=28\)

\(\Leftrightarrow\left(\dfrac{4}{DC}+\dfrac{8}{AB}\right)\left(DC+AB\right)=28\)

\(\Leftrightarrow4+8+\dfrac{4AB}{DC}+\dfrac{8DC}{AB}=28\)

\(\Leftrightarrow\dfrac{4AB}{DC}+\dfrac{8DC}{AB}=16\)

\(\Leftrightarrow\dfrac{AB}{DC}+\dfrac{2DC}{AB}=4\)

Đặt \(\dfrac{AB}{CD}=p\)

\(\Leftrightarrow p+\dfrac{2}{p}=4\)

=>  \(\left[{}\begin{matrix}p=2+\sqrt{2}\left(tm\right)\\p=2-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

         Vậy \(\dfrac{AB}{CD}=2+\sqrt[]{2}\)