In a circle of radius 5 cm, AB and AC are two chords such that  $\mathrm{AB}=\mathrm{AC}=6\mathrm{cm}$, then the length of the chord BC is

AO is internal bisector of $\angle \mathrm{BAC}\mathrm{\hspace{0.17em}}\left(\mathrm{AB}=\mathrm{AC}\right)$
$$\begin{gathered} \Rightarrow \frac{\mathrm{BM}}{\mathrm{CM}}=\frac{6}{6}=1\Rightarrow \mathrm{BM}=\mathrm{CM} \\ \mathrm{Let} \mathrm{BM}=\mathrm{x}\mathrm{cm} \\ \mathrm{AM}=\mathrm{y}\mathrm{cm} \\ \mathrm{OM}=\mathrm{OA}-\mathrm{AM}=\left(5-\mathrm{y}\right)\mathrm{cm} \end{gathered}$$
From right angled triangle AMB,
${\mathrm{x}}^2+{\mathrm{y}}^2=36\to \left(1\right)$
From right angled triangle OMB,
$$\begin{gathered} {\mathrm{x}}^2+{\left(5-\mathrm{y}\right)}^2=25\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-10\mathrm{y}=25-25 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2=10\mathrm{y}\to \left(2\right) \end{gathered}$$
From (1) and (2),
$$\begin{gathered} 10\mathrm{y}=36\Rightarrow \mathrm{y}=3.6\mathrm{cm} \\ \mathrm{From}\left(1\right), {\mathrm{x}}^2=36-{\left(3.6\right)}^2 \\ \mathrm{x}=\sqrt{36-{\left(3.6\right)}^2}=4.8\mathrm{cm} \\ \mathrm{BC}=2\mathrm{x}=2\times 4.8\mathrm{cm}=9.6 \mathrm{cm} \end{gathered}$$