If the median of a $ riangle A B C$ through $A$ is perpendicular to $A C$, then $\frac{ an A}{ an C}=$

Question

If the median of a $	riangle A B C$ through $A$ is perpendicular to $A C$, then $\frac{	an A}{	an C}=$

Aicharya · Accepted Answer

$A B C$ 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} $

If the median of a $\triangle A B C$ through $A$ is perpendicular to $A C$, then $\frac{\tan A}{\tan C}=$

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