First, notice that ATE~BKE, this gives [tex:fa9c62f2f7]\frac{EK}{x} = \frac{40}{EK}[/tex:fa9c62f2f7]. Then, by ATE~AKC, [tex:fa9c62f2f7]\frac{EK}{AE+EK}=\frac{x}{45}[/tex:fa9c62f2f7] Let [tex:fa9c62f2f7]EK=y[/tex:fa9c62f2f7]. Then, [tex:fa9c62f2f7]y^2=40x[/tex:fa9c62f2f7] and [tex:fa9c62f2f7]45y=x(y+\sqrt{x^2+y^2})[/tex:fa9c62f2f7] Substitute [tex:fa9c62f2f7]x=\frac{y^2}{40}[/tex:fa9c62f2f7] into the second one. [tex:fa9c62f2f7]45y=\frac{y^2}{40}(y+\sqrt{\frac{y^4}{1600}+y^2})[/tex:fa9c62f2f7] [tex:fa9c62f2f7]72000=y^2(40+\sqrt{y^2+1600})[/tex:fa9c62f2f7] Let [tex:fa9c62f2f7]z=\sqrt{y^2+1600}[/tex:fa9c62f2f7] [tex:fa9c62f2f7]72000=(z^2-1600)(40+z)[/tex:fa9c62f2f7]
Now, you can solve for z (s cubic equation) and then y and finally x.