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

You don't need that BC=AC, The simplest way to solve is using cosinus law: $(BD)^2=(AB)^2+(AD)^2-2(AB)(AD)cosÂ$ notice that $BD=AB$, so $AB/AD= 1/2cosÂ$.
Nonetheless, if you want to use similarity just realize that the triangles ABC and ABD are similar because they have same angles (this time you are using the fact BC=AC).
