Check the picture below, on the left side atop
AC is a diametrical line, and thus the angle that it makes is 180°, so the arcAB is obvious. Now, the angle ADB, well is an inscribed angle touching the arcAB.
Check the picture below, on the left side below
well, that triangle is a triangle made using the diametrical AC, and thus the angle ABC is an inscribed angle in a semicircle, now check the picture on the right side
when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one.
[tex]\cfrac{x}{6}~~ = ~~\cfrac{6}{13-x}\implies 13x-x^2=36\implies 13x=x^2+36 \\\\\\ 0=x^2-13x+36\implies 0=(x-9)(x-4)\implies x=AE= \begin{cases} 9~~\checkmark\\ 4 \end{cases}[/tex]
now, let's notice something, 4 is also a valid value, why we can't use 4?
well, let's notice, the segment AE is passed the half of AC, namely passed the center, since AC is 13, half that must be 6.5, so whatever AE is, it must be greater than 6.5, so the value 4 is no dice.
