## Online Course Discussion Forum

### MCIII Geometry Problem 5.8

MCIII Geometry Problem 5.8

Let $\angle BCD = \alpha$. From the trig form of Ceva's Theorem, I got $\frac{\sin{10^\circ}}{\cos{20^\circ}} \cdot \frac{\sin{(50-\alpha)}}{\sin{\alpha}} \cdot \frac{\sin{40^\circ}}{\sin{10^\circ}} = 1$. Applying the angle subtraction formula, I got $\frac{\cos{40^\circ}}{\tan{\alpha}} - \sin{40^\circ} = \frac{1}{2\sin{20^\circ}}$. I am not sure what to do next. Could I have a hint? Thank you.