Ratio of lengths in isosceles triangle 
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

Draw the perpendicular from $B$ to $AC$, meeting $AC$ at $X$. Then $\dfrac{AX}{AB}=\cos A$. But $AD=2AX$, and therefore $\dfrac{AB}{AD}=\dfrac{1}{2\cos A}$. 
