In a Circle, given segments $\overline{AC}$ = 13, $\overline{BE}$ = 6, $\overline{BE}$=$\overline{DE}$, find $\overline{AE}$ (E point where segments $\overline{AC}$ and $\overline{DB}$ cross)
It is known that when two chords intersect each other inside a circle, the products of their segments are equal. So: $\overline{DE}$ $\overline{EB}$ = $\overline{AE}$ $\overline{EC}$
Tried Pitagorian and other relations but could not find a way to get the result
So: $\;\overline{DE}$ $\overline{EB}$ = $\overline{AE}$ $\overline{EC}$
Hint: write it as $\;BE^2 = AE(AC-AE)\,$ and solve for $\,AE\,$.
$BE$ is and altitude of $\Delta ABC$ and we know that $\measuredangle ABC=90^{\circ}$.
Thus, $$BE^2=AE\cdot EC.$$ Let $AE=x$.
Thus, $x(13-x)=6^2,$ which says $x=4$ or $x=9$.
$$\overline { DE } \overline { EB } =\overline { AE } \overline { EC } \\ { 6 }^{ 2 }=\left( 13-x \right) x\\ { x }^{ 2 }-13x+36=0\\ \left( x-4 \right) \left( x-9 \right) =0\\ { x }_{ 1 }=4,{ x }_{ 2 }=9\\ \overline { AE } =9,\overline { EC } =4\\ $$
