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Triangle $ABC$ is isosceles with $AB = AC,$ and $D$ is the midpoint of $\overline{AB}.$ If $\angle BCD = \angle BAC = \theta,$ then find $\cos \theta.$

What I have done is I have set the AB and AC as length two, and utilized law of cosines to find the length of BC in terms of $\cos\theta$. After that I utilized law of cosines on triangle CBD to create an equation in which $\cos\theta$ could be found, but it turned out really complicated and I was unable to get anywhere. How do I solve this?

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$$\cos\theta = \frac{|\overline{AE}|}{|\overline{AC}|}= \frac{3}{4}$$

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Let $a=\frac{\overline{AB}}{2}$ and $b=\overline{BC}$. If you determine the angles in the ABC and CDB triangles you will find, that the two triangles are similar. From this you can find that $b^2=2 a^2$. Using this and the law of cosines for the CDB triangle you can find $\cos \theta$: $$ a^2=b^2+b^2-2b^2\cos\theta$$ $$ 3a^2=4a^2\cos\theta$$ $$ \cos\theta=\frac{3}{4}$$

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We see that $\measuredangle ACD=90^{\circ}-\frac{\theta}{2}-\theta=90^{\circ}-\frac{3\theta}{2}$ and $\measuredangle ADC=90^{\circ}-\frac{\theta}{2}+\theta=90^{\circ}+\frac{\theta}{2}$.

Thus, by law of sines for $\Delta ADC$ we obtain: $$\frac{AC}{\sin\left(90^{\circ}+\frac{\theta}{2}\right)}=\frac{AD}{\sin\left(90^{\circ}-\frac{3\theta}{2}\right)}$$ or $$\frac{2}{\cos\frac{\theta}{2}}=\frac{1}{\cos\frac{3\theta}{2}}$$ or $$2=\frac{1}{4\cos^2\frac{\theta}{2}-3}$$ or $$2=\frac{1}{2+2\cos\theta-3}$$ or $$\cos\theta=\frac{3}{4}.$$

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