Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.
1. Theorem: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.
According to this theorem,
[tex]BC^2=BD\cdot AB.[/tex]
Let BC=x cm, then AD=BC=x cm and BD=AB-AD=3-x cm. Then
[tex]x^2=(3-x)\cdot 3,\\ \\x^2=9-3x,\\ \\x^2+3x-9=0,\\ \\D=3^2-4\cdot (-9)=9+36=45,\\ \\\sqrt{D}=\sqrt{45}=3\sqrt{5},\\ \\x_1=\dfrac{-3-3\sqrt{5} }{2}<0,\ x_2=\dfrac{-3+3\sqrt{5} }{2}>0.[/tex]
Take positive value x. You get
[tex]AD=BC=\dfrac{-3+3\sqrt{5} }{2}\ cm.[/tex]
2. According to the previous theorem,
[tex]AC^2=AD\cdot AB.[/tex]
Then
[tex]AC^2=\dfrac{-3+3\sqrt{5} }{2}\cdot 3=\dfrac{-9+9\sqrt{5} }{2},\\ \\AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.[/tex]
Answer: [tex]AC=\sqrt{\dfrac{-9+9\sqrt{5} }{2}}\ cm.[/tex]
This solution doesn't need CD=2 cm. Note that if AB=3cm and CD=2cm, then
[tex]CD^2=AD\cdot DB,\\ \\2^2=AD\cdot (3-AD),\\ \\AD^2-3AD+4=0,\\ \\D<0[/tex]
This means that you cannot find solutions of this equation. Then CD≠2 cm.
