Suppose that a triangle ABC which is inscribed in a circle O of radius 2 satisfies

$$3\overrightarrow{\mathrm{OA}} + 4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}} = \vec{0}.$$

Let D denote the point of intersection of the straight line AO and the segment BC. We are to find the lengths of the segments AD and BD.

(1) When we set $\overrightarrow{\mathrm{OD}} = k\overrightarrow{\mathrm{OA}}$ where $k$ is a real number, we have

$$\overrightarrow{\mathrm{OD}} = -\frac{\mathbf{A}}{\mathbf{B}}k\overrightarrow{\mathrm{OB}} - \frac{\mathbf{C}}{\mathbf{D}}k\overrightarrow{\mathrm{OC}}.$$

As the three points B, C and D are located on a straight line, we obtain $k = \frac{\mathbf{EF}}{\mathbf{GH}}$.\\
From this we derive that $\mathrm{OD} = \mathbf{H}$ and finally obtain

$$\mathrm{AD} = 1.$$

(2) From (1) we see that $\mathrm{BD} = \frac{\mathbf{J}}{\mathbf{K}}\mathrm{BC}$. So in order to find the length of the segment BD, we should find the length of the segment BC.

First we note that

$$\mathrm{BC}^2 = \mathbf{L} - \mathbf{M}\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$$

where $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$ represents the inner product of $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$.\\
Since we know from (1) that $|4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}}|^2 = \mathbf{NO}$, we have

$$\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} = \frac{\mathbf{PQR}}{\mathbf{S}}.$$

Hence we obtain $\mathrm{BC} = \frac{\square\sqrt{\mathbf{U}}}{\square\mathbf{V}}$ and finally from that

$$\mathrm{BD} = \frac{\sqrt{\mathrm{W}}}{\mathrm{X}}.$$