Math, Please Help

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

$\text{Let $\angle DBC=\angle ABD = \alpha $}$

$1.\ \mathbf{ GP}:$

$\begin{array}{|rcll|} \hline \tan (\angle CBE) &=& a \\ \tan (\angle DBE) &=& ar \\ \tan (\angle ABE) &=& ar^2 \\\\ a\cdot ar^2 &=& (ar)^2 \\ \tan (\angle CBE)\cdot \tan (\angle ABE) &=& \tan^2(\angle DBE) \quad | \quad \angle CBE=\angle DBE-\alpha,\ \angle ABE=\angle DBE+\alpha\\ \tan (\angle DBE-\alpha)\cdot \tan (\angle DBE+\alpha) &=& \tan^2(\angle DBE) \\ && \boxed{ \text{Formula}\\ \tan(x+y)=\dfrac{\tan(x)+\tan(y)} {1-\tan(x)\tan(y)}\\ \tan(x-y)=\dfrac{\tan(x)-\tan(y)} {1+\tan(x)\tan(y)} \\ \tan(x-y)\tan(x+y)= \left(\dfrac{\tan(x)-\tan(y)} {1+\tan(x)\tan(y)}\right) \left(\dfrac{\tan(x)+\tan(y)} {1-\tan(x)\tan(y)}\right) \\ =\dfrac{\tan^2(x)-\tan^2(y)}{1-\tan^2(x)\tan^2(y)} } \\\\ \dfrac{\tan^2(\angle DBE)-\tan^2(\alpha)} {1-\tan^2(\angle DBE)\tan^2(\alpha)} &=& \tan^2(\angle DBE) \\ \tan^2(\angle DBE)-\tan^2(\alpha) &=& \left(1-\tan^2(\angle DBE)\tan^2(\alpha) \right)\tan^2(\angle DBE) \\ -\tan^2(\alpha) &=& -\tan^4(\angle DBE)\tan^2(\alpha) \\ \tan^4(\angle DBE)\tan^2(\alpha) -\tan^2(\alpha) &=& 0 \\ \underbrace{ \tan^2(\alpha)}_{\neq 0}\underbrace{(\tan^4(\angle DBE)-1)}_{=0} &=& 0 \\ \tan^4(\angle DBE)-1 &=& 0 \\ \tan^4(\angle DBE) &=& 1 \\ \tan (\angle DBE) &=& 1 \\ \angle DBE &=& \arctan(1) \\ \mathbf{\angle DBE} &=& \mathbf{45^\circ} \\ \hline \end{array}$

$2.\ \mathbf{ AP}:$

$\begin{array}{|rcll|} \hline \cot (\angle DBE) &=& b \quad | \quad \angle DBE = 45^\circ \\ \cot (45^\circ) &=& b \\ 1 &=& b \\\\ \cot (\angle CBE)=\cot (\angle DBE-\alpha)=\cot (45^\circ-\alpha) = b+d&=&1+b \\ \cot (\angle DBC)=\cot(\alpha) = b+2d=1+2d \\\\ \cot (\angle DBE) &=& 1 \\ \cot (45^\circ-\alpha) &=& 1+d \\ \cot (\alpha) &=& 1+2d \\\\ 2(1+d) &=& 1+(1+2d) \\ 2\cot (\angle 45^\circ-\alpha) &=& 1+ \cot (\alpha) \\ && \boxed{ \text{Formula}\\ \cot(x-y)=\dfrac{\cot(x)\cot(y)+1}{\cot(y)-\cot(x)}\\ \cot(45^\circ-\alpha)=\dfrac{\cot(45^\circ)\cot(\alpha)+1}{\cot(\alpha)-\cot(45^\circ)}\\ =\cot(45^\circ-\alpha)=\dfrac{1*\cot(\alpha)+1}{\cot(\alpha)-1}\\ =\cot(45^\circ-\alpha)=\dfrac{\cot(\alpha)+1}{\cot(\alpha)-1} } \\\\ 2\left(\dfrac{\cot(\alpha)+1}{\cot(\alpha)-1}\right) &=& 1+ \cot (\alpha) \\ (1+ \cot (\alpha))(\cot(\alpha)-1)&=& 2 (\cot(\alpha)+1)\\ (1+ \cot (\alpha))((\cot(\alpha)-1)-2 (\cot(\alpha)+1)&=& 0\\ (1+ \underbrace{\cot(\alpha)}_{\neq-1} ) ( \underbrace{\cot(\alpha)-3}_{=0} ) &=& 0\\ \cot(\alpha)-3 &=& 0 \\ \mathbf{\cot(\alpha)} &=& \mathbf{3} \\ \hline \end{array}$

$\mathbf{BD=\ ?}$

$\begin{array}{|rcll|} \hline \cos(\angle DBE) &=& \dfrac{BD}{BE} \\ \cos(45^\circ) &=& \dfrac{BD}{10} \\ BD &=& 10\cos(45^\circ) \\ BD &=& 10\dfrac{\sqrt{2}}{2} \\ \mathbf{BD} &=& \mathbf{5 \sqrt{2} } \\ \hline \end{array}$

$\mathbf{AC=\ ?}$

$\begin{array}{|rcll|} \hline \cot(\alpha)&=& \dfrac{BD}{DC}\times \dfrac{2}{2} \\\\ \cot(\alpha)&=& \dfrac{2BD}{2DC} \quad | \quad 2DC = AC \\\\ \cot(\alpha)&=& \dfrac{2BD}{AC} \\\\ AC&=& \dfrac{2BD}{\cot(\alpha)} \quad | \quad \cot(\alpha) = 3 \\\\ AC&=& \dfrac{2BD}{3} \quad | \quad \mathbf{BD = 5 \sqrt{2} }\\ AC&=& \dfrac{2 *5 \sqrt{2}}{3} \\ \mathbf{AC} &=& \mathbf{\dfrac{10 \sqrt{2}}{3}} \\ \hline \end{array}$

$\mathbf{\text{area of $\triangle ABC$ }}$

$\begin{array}{|rcll|} \hline \text{area of $\triangle ABC$ } &=& \dfrac{AC\cdot DB}{2} \\ \text{area of $\triangle ABC$ } &=& \dfrac{\dfrac{10 \sqrt{2}}{3}\cdot 5 \sqrt{2}}{2} \\ \mathbf{\text{area of $\triangle ABC$ }} &=& \mathbf{\dfrac{50}{3}} \\ \hline \end{array}$

source: https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_24