\( \eqalign{ & \frac{{BC}} {{\sin 20^\circ }} = \frac{{AC}} {{\sin 80^\circ }} \Rightarrow BC = \frac{{AC \cdot \sin 20^\circ }} {{\sin 80^\circ }} \cr & \frac{{CD}} {{\sin 40^\circ }} = \frac{{AC}} {{\sin 100^\circ }} \Rightarrow CD = \frac{{AC \cdot \sin 40^\circ }} {{\sin 100^\circ }} \cr & BC + CD = \frac{{AC \cdot \sin 20^\circ }} {{\sin 80^\circ }} + \frac{{AC \cdot \sin 40^\circ }} {{\sin 100^\circ }} \cr & BC + CD = \frac{{AC \cdot \sin 20^\circ }} {{\sin 80^\circ }} + \frac{{AC \cdot \sin 40^\circ }} {{\sin 80^\circ }} \cr & BC + CD = \frac{{AC \cdot \sin 20^\circ + AC \cdot \sin 40^\circ }} {{\sin 80^\circ }} \cr & BC + CD = \frac{{AC \cdot \left( {\sin 20^\circ + \sin 40^\circ } \right)}} {{\sin 80^\circ }} \cr & BC + CD = \frac{{AC \cdot \left( {2\sin \frac{{20^\circ + 40^\circ }} {2}\cos \frac{{20^\circ - 40^\circ }} {2}} \right)}} {{\sin 80^\circ }} \cr & BC + CD = \frac{{AC \cdot \left( {2\sin 30^\circ \cos 10^\circ } \right)}} {{\sin 80^\circ }} \cr & BC + CD = \frac{{AC \cdot \cos 10^\circ }} {{\sin 80^\circ }} = AC \cr } \)
Omdat het kan...:-)
