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Given, \(AD=5, AE=4\) and \(\angle DBC=90^\circ \). As \(D,B,C\) are points on the circle having radius 6. Therefore \(DC\) is the diameter of the circle \(\Rightarrow DC=6×2=12\).

Now DC is diameter of the circle \(\Rightarrow \angle DEC=90^\circ \). Therefore \(\angle DEA\) is also right angle.

The length of \(DE=\sqrt{5^2-4^2}=3\). And the length of \(EC=\sqrt{12^2-3^2}=3\sqrt{15}\). Therefore \(AC=AE+EC=4+3\sqrt{15}\).

From \(∆AED\) and \(∆ABC\) we have, \(\frac{AD}{DE}=\frac{AC}{BC} \Rightarrow BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}\). Therefore the length of \(BC\) is \(\frac{12+9\sqrt{15}}{5}\). (Ans.)

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