$A B C$ is the triangle while $A M$ is the median $A C$ and $A M$ are perpendicular.
$$
\Rightarrow \quad \angle M A C=90^{\circ}
$$
Since, $A M$ is the median, $M$ is the mid-point of line $B C$.
$$
\Rightarrow \quad B M=C M=2 B C
$$
Draw a line perpendicular to $A M$ (through $M$ ) let it intersect the line $A B$ at $P$.
In $\triangle A B C$,
$$
\angle A+\angle B+\angle C=180^{\circ}
$$
In $\triangle A M C$,
$$
\begin{array}{rlrl}
& \angle A M C+\angle C A M+\angle M C A & =180^{\circ} \\
\Rightarrow & & \angle A M C+90^{\circ}+\angle C & =180^{\circ} \\
\Rightarrow & \angle A M C & =90^{\circ}-\angle C
\end{array}
$$
Again at point $M$.
$$
\begin{array}{rlrl}
& & \angle B M P+\angle P M A+\angle A M C & =180^{\circ} \\
\Rightarrow & \angle B M P+90^{\circ}+90^{\circ}-\angle C & =180^{\circ} \\
\Rightarrow & & \angle B M P & =\angle C
\end{array}
$$
But in $\triangle B P M$,
$$
\begin{aligned}
& \angle B P M+\angle B M P+\angle P B M=180^{\circ} \\
& \Rightarrow \quad \angle B P M+\angle C+\angle B=180^{\circ} \\
& \angle B P M=\angle A \\
&
\end{aligned}
$$
[From Eqs. (i) and (iv)]
Now, it is evident that $A B C$ and $M P B$ are equivalent triangles.
$\Rightarrow \quad \frac{B M}{B C}=\frac{B P}{A B}=\frac{P M}{A C}=\frac{1}{2}$
Also,
$$
\begin{aligned}
& \angle B A C=\angle B A M+\angle M A C \\
& \Rightarrow \quad \angle B A M=\angle B A C-\angle M A C \\
& \Rightarrow \quad \angle B A M=\angle A-90^{\circ} \\
&
\end{aligned}
$$
In $\triangle A P M$,
$$
\begin{array}{rlrl}
& & \angle P A M+\angle A P M+\angle A M P & =180^{\circ} \\
\Rightarrow & & \angle A-90^{\circ}+\angle A P M+90^{\circ}=180^{\circ} \\
\Rightarrow & & \angle A P M & =180^{\circ}-A \\
& & \tan (\angle A P M) & =\frac{A M}{P M} \\
\Rightarrow & & \tan \left(180^{\circ}-A\right)=\frac{A M}{P M} \\
\Rightarrow & & -\tan A=\frac{A M}{P M} \\
\Rightarrow & & A M & =-P M \tan A
\end{array}
$$
Now, in $\triangle A C M, \tan C=\frac{A M}{A C}$
$$
\Rightarrow \quad A M=A C \tan C
$$
[From Eqs. (viii) and (ix),]
$$
A M=A C \tan C=-P M \tan A
$$
$\Rightarrow P M \tan A+A C \tan C=0$
$$
\begin{aligned}
\Rightarrow & \frac{M P}{A C}=\frac{1}{2} \\
\Rightarrow & A C=2 M P
\end{aligned}
$$
Hence, $P M \tan A+2 P M \tan C=0$
$$
\begin{array}{rlrl}
\Rightarrow & P M(\tan A+2 \tan C) & =0 \\
\Rightarrow & & \tan A+2 \tan C & =0 \\
\Rightarrow & & \tan A & =-2 \tan C \\
\Rightarrow & & \frac{\tan A}{\tan C} & =-2
\end{array}
$$