$S=\frac 12.bc\sin A=\frac 12.AE.a\ ,\frac{a}{\sin A}=2R$then $bc=AE.2R$

Let $w=C(O,R)$ be the circumcircle of the triangle $ABC$. Define the point $A'\in w\cap (AO$. Then $\widehat {ABE}\equiv\widehat {AA'C}\ ,$ $AE\perp BE$ and $AC\perp A'C$ $\Longrightarrow$ $\triangle ABE\sim \triangle AA'C$ $\Longrightarrow \frac{AB}{AA'}=\frac{BE}{A'C}=\frac{EA}{CA}$ $\Longrightarrow$ $\frac{c}{2R}=\frac{h_a}{b}$ $\Longrightarrow$ $\boxed {\ bc=2Rh_a\ }\ .$Remark. $S=\frac 12ah_a=$ $\frac 12 a\cdot \frac{bc}{2R}$ $\Longrightarrow$ $\boxed {\ 4RS=abc\ }\ .$

The problem tells us to prove the following well-known relation in a trinagle i.e \[ h_{a}=\frac{bc}{2R}, \] where $h_{a}$ is the altitude of the triangle $ABC$ and $a, b, c$ are the side lenghts of the same triangle, $ABC$. Well, here is one-lin solution: We know that \[ S=\frac{abc}{4R} \] and \[ a\cdot h_{a}=2S. \] The conclusion follows immediately.