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This question already has an answer here:

I saw the following problem on Facebook (figure not drawn to scale):

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$\triangle ABC$ is an isosceles triangle with $AB\equiv BC$, $AC\equiv BD$, and $\angle B=20^\circ$. Find $\angle CAD$.

What I did was let $BC\equiv x$ and divide the triangle in two by bisecting $\angle B$. This yielded two right triangles with base $x\cos80^\circ$. Therefore, the base of $\triangle ABC$ has length $2x\cos80^\circ$.

Clearly, $DC=x-2x\cos80^\circ$, and $\angle C=80^\circ$.

Now, to find $AD\equiv y$, I used the law of cosines:

$$ y=\sqrt{(2x\cos80^\circ)^2+(x-2x\cos80^\circ)^2-2(2x\cos80^\circ)(x-2x\cos80^\circ)\cos80^\circ}. $$

Finally, to find $\angle CAD\equiv\theta$, I used the law of sines:

$$\begin{align} \frac{\sin80^\circ}{y}&=\frac{\sin\theta}{x-2x\cos80^\circ}\\ \Longrightarrow\theta&=70^\circ. \end{align}$$

Is my result correct? Is there an easier way to solve this problem?

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marked as duplicate by azimut, Thomas, Daniel Robert-Nicoud, Nick Peterson, Vedran Šego Oct 29 '13 at 18:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ While using trigonometry does give an answer, these kind of question usually requires an answer using geometry only. See cut-the-knot.org/triangle/80-80-20/IndexToLong.shtml $\endgroup$ – peterwhy Oct 29 '13 at 17:01
  • $\begingroup$ I see. Thank you for the link! $\endgroup$ – wjm Oct 29 '13 at 17:06
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    $\begingroup$ For example, construct $E$ in the triangle such that $ACE$ is equilateral. Then $BE$ bisects angle $B$ and triangle $DBA$ is congruent to triangle $ECB$. Thus $\angle{CAD}=80^{\circ}-\angle{BAD}=80^{\circ}-\angle{CBE}=80^{\circ}- \frac{\angle{CBA}}{2}=70^{\circ}$ $\endgroup$ – Ivan Loh Oct 29 '13 at 17:07
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    $\begingroup$ @peterwhy Interestingly enough, all seven solutions at the link you gave are different from the one I came up with. $\endgroup$ – Ivan Loh Oct 29 '13 at 17:14