In $\triangle ABC$, $\ AB=AC$, $\ BD=DC$, and $\ BE>CE\ $. If $\ \tan\angle EAC$, $\ \tan\angle EAD$, $\ \tan\angle EAB$ are in geometric progression, and $\ \cot\angle DAE$, $\ \cot\angle CAE$, $\ \cot\angle DAB$ are in arithmetic progression, and $AE=10$, then what is the area of the $\triangle ABC$?
(The answer is $[ABC]=50/3$.)
I called $\angle EAC=\alpha$, $\angle CAD=\angle DAB=\beta$, and, using the AP and GP properties and the relations of $\tan(\alpha+\beta)$ and $\tan(2\beta)$, I wrote the system:
$$\begin{cases}x^2=y^2(2+x)\\[4pt] \left(1-y(y+2x)\right)(x+y)^2=(1-xy)(y^2+1)x \end{cases}$$
where $x=\tan\alpha$ and $y=\tan\beta$.
Βut I can't solve this system.
How can I solve this problem in a better way?
Thanks for attention.
The given geometric and arithmetic series leads to
$$\tan^2\alpha = \tan(\alpha-\beta)\tan(\alpha+\beta),\>\>\>\>\> 2\cot(\alpha-\beta) = \cot\alpha+\cot\beta$$
Apply the identities $\tan(a\pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b}$ to get
$$\tan^2\beta(1-\tan^4\alpha)=0, \>\>\>\>\> 2(1+\tan\alpha\tan\beta)=\frac{\tan\alpha}{\tan\beta} - \frac{\tan\beta}{\tan\alpha}$$
Solve to obtain the valid solution $\tan\alpha = 1$ and $\tan\beta = \frac13$. Then, the area of the triangle ABC is
$$Area = AD\cdot BD = AE^2 \cos^2\alpha\tan\beta = 10^2\left(\frac1{\sqrt2}\right)^2\frac13=\frac{50}3$$