<s>[INST] What's the measure of the segment $EF$ in the triangle below? For reference:

The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)

My progress:

Here are the relations I found:
$EF = R - OE \\
\triangle AEF: EF^2 + OE^2 = AF^2\implies\\
(R-OE)^2 + (\frac{9}{2})^2 = AF^2 \implies:
(R-OE)^2 + \frac{81}{4} = AF^2\\
\triangle AOE: OE^2+AE^2 = AO^2 \implies OE^2 +(\frac{9}{2})^2 =R^2 \implies OE^2+ \frac{81}{4} = R^2\\
 \triangle AOF: AF^2 = R^2 +OF^2-2OF.OE \implies\\
AF^2 = R^2 +R^2-2R.OE \therefore AF^2 = 2R^2-2R.OE = 2R\underbrace{(R-OE)}_{=EF}$
...??
 [/INST] Draw altitude $AD$ and call $BD=x$.

From Pythagoras' theorem, $$7^2-x^2=9^2-(8-x)^2\implies x=2$$
It follows that $AD=3\sqrt5$.
We see $\triangle ABD\sim\triangle AOE$. Therefore, $$\frac{AO}7=\frac{OE}2=\frac{9/2}{3\sqrt5}\\\implies AO=\frac{3\cdot7}{2\sqrt5},\: OE=\frac{3\cdot2}{2\sqrt5}$$
Thus $$EF=\underbrace{OF}_{=AO}-OE=\frac{3\cdot5}{2\sqrt5}=\frac{3\sqrt5}{2}$$
</s>