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

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

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

*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 $EF$
A: 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 $$EF=\underbrace{OF}_{=AO}-OE=\frac{3\cdot5}{2\sqrt5}=\frac{3\sqrt5}{2}$$
A: 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}$
