O $\triangle ABC$ is inscribed in a circle with center $O$. Through $B$ we draw $BD ⊥ AO$, which meets $AC$ in $D$. A$B = 6$ and $AC = 9$. Calculate $AD$. (S:$AD=4$)
I try
$2R.BC =BG.AC+AB.CG \implies 2R.BC = 9BG+6CG$
$\triangle ABG: BG^2 = 4R^2-36$
$\triangle ACG: CG^2 = 4R^2 - 81$
The angle $\widehat{OAB}=\widehat{OBA}$ is $\frac 12(180^\circ-2\hat C)=90^\circ-\hat C$. So $\widehat{ABD}=\hat C$, making $\widehat{ADB}=\hat B$. The two triangles are thus similar: $$\Delta ABC\sim\Delta ADB\qquad \text{ which gives } \frac{AB}{AD} = \frac{AC}{AB} \ . $$ The missing segment is the yellow one in the equality: $6^2=9\cdot \bbox[yellow]4$.