A geometry problem by A Former Brilliant Member

$TanA=Tan45=1.\ \ \ Let\ \ TanB=1+d,\ \ \ and\ \ \ TanC=1+2d.\\ But\ in\ a\ \triangle\ TanA+TanB+TanC=TanA*TanB*TanC.\\ \therefore\ \ 1+(1+d)+(1+2d) = 1*(1+d)*(1+2d)\\ \implies\ 3(1+d)=(1+d)(1+2d),\ \ Since\ B\neq 0,\ TanB\neq 0,\ \therefore\ 1+d\neq 0.\\ So\ 3=1+2d,\ \ \implies\ d=1.\\ So\ A=Tan^{-1}1,\ \ \ B=Tan^{-1}2\ \ \ C=Tan^{-1}3.\\ \therefore\ TanC=\Large\ \ \color{#D61F06}{3}.\\\ \ \ \\$

$\angle{A}=45^{\circ}$ . So $\tan{A}=1$ . Let $\begin{aligned}1=\tan{B}-d\end{aligned}$ and $\begin{aligned}\tan{C}=\tan{B}+{d} \end{aligned}$ ; where $\begin{aligned}\text{d is common difference}\end{aligned}$

Using the triangle identity $\begin{aligned}\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}\end{aligned}$ , we get $\begin{aligned}3{\tan{B}}=\tan{B}\tan{C}\end{aligned}$ . And hence $\begin{aligned}\tan{C}=3\end{aligned}$

Find 2tan(B)=1+tan(C)

Use tan(A+C)= (1+tan(C))/(1-tan(C))

Find tan(C)=3