<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] *

*$\angle ABC=\frac12\angle AOC=\angle AOE$

*$\sin B=\sin\angle AOE=\frac{AC/2}{R}=\frac{b}{2R}$

*Similarly
$$2R=\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$$

*Area of $\triangle ABC$ is $$K=\frac12ab\sin C=\frac{abc}{4R}$$

*Use Heron's formula to calculate area $K$

*Get $R$

*Use Pythagoras to get $OE$

*Get $EF$
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