# 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

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Please post the text of the question in the body and not in the title. Also, this looks like a homework problem, so I've added the homework tag. If not, you can remove the tag. –  Rahul Nov 16 '12 at 6:29

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

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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).

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The two answers provided by Rafael and André are correct, $\frac{AB}{AD}=\frac{1}{2}sec(\alpha)$, however I've added the following diagram for illustration purposes, in case the wording of the problem was somehow incorrect, as the answer provided by OP does not seem consistent with the description of the problem as its understood here.

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