In $\triangle ABC$ , $BC = AC$. Also $D$ is a point on side $AC$ such that $BD = AB$. Find the ratio $\frac{AB}{AD}$. Justify your answer.
The answer is supposed to be $\frac1 {cosA}$ where $A = \angle BAC$. I can't figure out how to get there: Related Topics: Similarity, Areas, Golden Ratio