Properties and Solution of Triangles
7.0 $m-n$ Rule
7.0 $m-n$ Rule
If in a triangle $ABC$, $D$ is the point on $BC$ such that $BD:DC = m:n$ and $\angle ADC = \theta ,\angle BAD = \alpha ,\angle DAC = \beta $, then $$\begin{equation} \begin{aligned} (m + n)\cot \theta = m\cot \alpha - n\cot \beta \\ (m + n)\cot \theta = n\cot B - m\cot C \\\end{aligned} \end{equation} $$
Proof: Consider $\Delta ABD$,
Using exterior angle theorem, we can say that $$\begin{equation} \begin{aligned} \theta = \alpha + \angle ABD \\ \angle ABD = \theta - \alpha \\\end{aligned} \end{equation} $$
Now using sine rule in $\Delta ABD$, we get
$$\frac{{\sin \alpha }}{{BD}} = \frac{{\sin (\theta - \alpha )}}{{AD}}$$
In $\Delta ADC$
$$\begin{equation} \begin{aligned} \frac{{\sin \beta }}{{DC}} = \frac{{\sin (\theta + \beta )}}{{AD}} \\ \angle C = \pi - \beta - \theta \\ \sin C = \sin (\pi - \beta - \theta ) \\ \sin C = \sin (\beta + \theta ) \\ \\\end{aligned} \end{equation} $$
Dividing both equations, we get $$\frac{{\sin \alpha }}{{BD}} \times \frac{{DC}}{{\sin \beta }} = \frac{{\sin (\theta - \alpha )}}{{\sin (\theta + \beta )}}$$
As $\frac{{DC}}{{BD}} = \frac{n}{m}$
$$\frac{n}{m}\frac{{\sin \alpha }}{{\sin \beta }} = \frac{{\sin \theta \cos \alpha - \cos \theta \sin \alpha }}{{\sin \theta \cos \beta + \cos \theta \sin \beta }}$$
By cross multiplication, we get
$$n\sin \alpha \sin \theta \cos \beta + n\sin \alpha \cos \theta \sin \beta = m\sin \beta \sin \theta \cos \alpha - m\sin \beta \cos \theta \sin \alpha $$
Dividing throughout by $\sin \theta \sin \alpha \sin \beta $, we get
$$\begin{equation} \begin{aligned} n\cot \beta + n\cot \theta = m\cot \alpha - m\cot \theta \\ (m + n)\cot \theta = m\cot \alpha - n\cot \beta \\\end{aligned} \end{equation} $$
Question 14. The base of a triangle is divided into three equal parts. If ${t_1},{t_2},{t_3}$ be the tangents of the angles subtended by these parts at opposite vertex, prove that $$4(1 + \frac{1}{{{t_1}^2}}) = (\frac{1}{{{t_1}}} + \frac{1}{{{t_2}}})(\frac{1}{{{t_2}}} + \frac{1}{{{t_3}}})$$
Solution: Let point $D$ and $E$ divides the base $BC$ into three equal parts i.e $BD=DE=EC=d$. Let $\alpha ,\beta ,\gamma $ are the angles subtended by $BD$, $DE$ and $EC$ respectively at their opposite vertex and $\angle AEC = \theta $. We have
$$\begin{equation} \begin{aligned} {t_1} = \tan \alpha \\ {t_2} = \tan \beta , \\ {t_3} = \tan \gamma \\\end{aligned} \end{equation} $$
Now in triangle $ABC$,
$$BE:EC = 2d:d = 2:1$$
From $m-n$ rule, we get
$$\begin{equation} \begin{aligned} (2 + 1)\cot \theta = 2\cot (\alpha + \beta ) - \cot \lambda \\ 3\cot \theta = 2\cot (\alpha + \beta ) - \cot \lambda ...(i) \\\end{aligned} \end{equation} $$
In triangle $ADC$, $$DE:EC = d:d = 1:1$$
If we apply $m-n$ rule in triangle $ADC$, we get
$$\begin{equation} \begin{aligned} (1 + 1)\cot \theta = 1.\cot \beta - 1.\cot \lambda \\ 2\cot \theta = \cot \beta - \cot \lambda ...(ii) \\\end{aligned} \end{equation} $$
From $(i)$ and $(ii)$, we get $$\frac{{3\cot \theta }}{{2\cot \theta }} = \frac{{2\cot (\alpha + \beta ) - \cot \lambda }}{{\cot \beta - \cot \lambda }}$$
by cross multiplication,
$$\begin{equation} \begin{aligned} 3\cot \beta - 3\cot \lambda = 4\cot (\alpha + \beta ) - 2\cot \lambda \\ 3\cot \beta - \cot \lambda = 4\cot (\alpha + \beta ) \\ 3\cot \beta - \cot \lambda = 4(\frac{{\cot \alpha \cot \beta - 1)}}{{\cot \alpha + \cot \beta )}} \\\end{aligned} \end{equation} $$ using $$\cot (\alpha + \beta ) = \frac{{\cot \alpha \cot \beta - 1}}{{\cot \alpha + \cot \beta }}$$
By cross multiplication, we get
$$\begin{equation} \begin{aligned} 3\cot \beta \cot \alpha + 3{\cot ^2}\beta - \cot \lambda \cot \alpha - \cot \lambda \cot \beta = 4\cot \alpha \cot \beta - 4 \\ 4 + 3{\cot ^2}\beta = \cot \alpha \cot \beta + \cot \lambda \cot \alpha + \cot \lambda \cot \beta \\ 4 + 4{\cot ^2}\beta = \cot \alpha \cot \beta + \cot \lambda \cot \alpha + \cot \lambda \cot \beta + {\cot ^2}\beta \\ 4(1 + {\cot ^2}\beta ) = (\cot \alpha + \cot \beta )(\cot \beta + \cot \lambda ) \\ 4(1 + \frac{1}{{{{\tan }^2}\beta }}) = (\frac{1}{{\tan \alpha }} + \frac{1}{{\tan \beta }})(\frac{1}{{\tan \beta }} + \frac{1}{{\tan \lambda }}) \\\end{aligned} \end{equation} $$
$$4(1 + \frac{1}{{{t_1}^2}}) = (\frac{1}{{{t_1}}} + \frac{1}{{{t_2}}})(\frac{1}{{{t_2}}} + \frac{1}{{{t_3}}})$$
