# Area of an isosceles triangle where the tangents of some angles are in geometric progression

In $\triangle ABC$, $AB=BC$ and $\overline{BD}$ is an altitude. Point $E$ is on the extension of $\overline{AC}$ such that $\overline{BE}=10$. The values of $\tan CBE$, $\tan DBE$, and $\tan ABE$ form a geometric progression, and the values of $\cot DBE$, $\cot CBE$, and $\cot DBC$ form an arithmetic progression. What is the area of $\triangle ABC$?

I'm pretty stuck on this problem. My first thought was to convert cotangent into tangent, but that got messy very quickly. To find the area of $\triangle ABC$, we would need to use the base and the height, or $BD$ and $AE$. I don't know how to go about finding those lengths. Any helps would be great! You know that $\triangle ABC$ is isosceles, and $D$ is the midpoint of the base across from $B$. Further, $\triangle BDE$ is a right triangle.
Let $\theta = \angle DBE$ and $\phi = \angle DBC$, so $\tan \angle CBE = \tan (\theta - \phi)$ and $\tan \angle ABE = \tan(\theta+\phi)$. Applying sum and difference formulas, we get $$\tan \angle CBE = \frac{\tan\theta-\tan \phi}{1+\tan\theta\tan \phi} = \frac{x-y}{1+xy}$$ $$\tan \angle ABE = \frac{\tan\theta+\tan \phi}{1-\tan\theta\tan \phi} = \frac{x+y}{1-xy}$$ where $x=\tan\theta$ and $y=\tan \phi$. For a geometric progression, we need $$\frac{\frac{x-y}{1+xy} }{x} = \frac{x}{ \frac{x+y}{1-xy} }$$ Simplifying, $$x^2 = \frac{x^2-y^2}{1-x^2y^2}$$ $$x^4 = 1$$ So the only valid solution is $x=1$, or $\angle DBE = \frac{\pi}{4}$.
Looking to the cotangents, $\cot\angle DBE = 1$, $\cot\angle CBE = \frac{1+y}{1-y}$, and $\cot\angle DBC = \frac{1}{y}$ must form an arithmetic progression, so $$\frac{1+y}{1-y} - 1 = \frac{1}{y} - \frac{1+y}{1-y}$$ Simplifying, $$3y^2-2y-1=0$$ So $y = \frac{1}{3},-1$. Clearly, $\frac{1}{3}$ is the only solution that makes sense. Now, the area of $\triangle ABC = \overline{BD}^2 \tan\angle DBC = \left(\frac{10}{\sqrt{2}}\right)^2 y = \frac{50}{3}$.