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:
$EF = R - OE \\
\triangle AEF: EF^2 + OE^2 = AF^2\implies\\
(R-OE)^2 + (\frac{9}{2})^2 = AF^2 \implies:
(R-OE)^2 + \frac{81}{4} = AF^2\\
\triangle AOE: OE^2+AE^2 = AO^2 \implies OE^2 +(\frac{9}{2})^2 =R^2 \implies OE^2+ \frac{81}{4} = R^2\\
 \triangle AOF: AF^2 = R^2 +OF^2-2OF.OE \implies\\
AF^2 = R^2 +R^2-2R.OE \therefore AF^2 = 2R^2-2R.OE = 2R\underbrace{(R-OE)}_{=EF}$
...??

Here is another approach -

Using Heron's formula, $\triangle_{ABC} = 12 \sqrt5$
$\triangle_{ABC} = \frac12 \cdot BH \cdot AC = 12 \sqrt5 \implies BH = \frac{8 \sqrt5}{3}$
Using Pythagoras, $ \displaystyle AH = \frac{11}{3} \implies HE = \frac92 - \frac{11}3 = \frac56$
As $BG$ is angle bisector, $ \displaystyle AG = \frac{7}{7+8} \cdot 9 = \frac{21}5 \implies GE = \frac3{10}$
That leads to $ \displaystyle HG = \frac8{15}$
As $ \triangle BHG \sim \triangle FEG, ~  \displaystyle \frac{EF}{BH} = \frac{GE}{HG} \implies EF = \frac{3 \sqrt5}{2}$
