Simple question, but I can't figure it out. Consider an isosceles triangle ABC with \alpha=\beta dissected by a line through C and D, where D is on AB. It is obvious that |CD|<=|AC|=|BC|, but I want to prove it using trigonometry. I can use |BD|<=|BC| in my assumptions but not angle(BCD)<=angle(ACB). Otherwise I'd be done, for the law of sines using the triangle ABC and BCD gives me |BC|>=|CD|*(sin(angle(BCD))/sin(angle(ACB))). But how to show that angle(BCD)<=angle(ACB), obviously without using what we are trying to prove, i.e. |CD|<=|BC|.