Consider a triangle ABC and its circumscribed circle O, where the lengths of the three sides of the triangle are $$\mathrm{AB} = 2, \quad \mathrm{BC} = 3, \quad \mathrm{CA} = 4.$$ Below, the area of a triangle such as PQR is expressed as $\triangle\mathrm{PQR}$.
(1) We see that $\cos\angle\mathrm{ABC} = \frac{\mathbf{AB}}{\mathbf{C}}$.
(2) Let us take a point D on the circumference of circle O such that it is on the opposite side of the circle from point B with respect to AC and $$\frac{\triangle\mathrm{ABD}}{\triangle\mathrm{BCD}} = \frac{8}{15}.$$ We are to find the lengths of line segments AD and CD.
First, since $$\angle\mathrm{BAD} = \mathbf{DEF}^\circ - \angle\mathrm{BCD},$$ we have $\sin\angle\mathrm{BAD} = \sin\angle\mathrm{BCD}$. Hence from (1) we have $$\frac{\mathrm{AD}}{\mathrm{CD}} = \frac{\mathbf{G}}{\mathbf{H}},$$ so we set $\mathrm{AD} = \mathbf{G}k$ and $\mathrm{CD} = \mathbf{H}k$, where $k$ is a positive number. Furthermore, since $$\angle\mathrm{ADC} = \mathbf{IJK}^\circ - \angle\mathrm{ABC},$$ we have $\cos\angle\mathrm{ADC} = \frac{\mathbf{L}}{\mathbf{L}}$. Hence, we obtain $k = \frac{\mathbf{N}}{\sqrt{\mathbf{OP}}}$, and then $$\mathrm{AD} = \frac{\mathbf{QR}\sqrt{\mathbf{OP}}}{\mathbf{OP}},$$ $$\mathrm{CD} = \frac{\mathbf{ST}\sqrt{\mathbf{OP}}}{\mathbf{OP}}.$$
(3) When we denote the point of intersection of the straight line DA and the straight line CB by E, we have $$\frac{\triangle\mathrm{ABE}}{\triangle\mathrm{CDE}} = \frac{\mathbf{UV}}{\mathbf{WXY}}.$$