# Properties of Triangle m – n Theorem

## Properties of Triangle m – n Theorem

m – n Theorem:

Let D be a point on the side BC of Triangle ABC such that BD. DC = m: n and angle BAD = α, angle ADC = θ and angle DAC = β then (m + n) cot θ = n cot B – m cot C

Proof:

BD/DC = m/n

From ΔABD,

$$\angle ADB=\left({{180}^{o}}-\theta \right)$$,

$$\angle BAD=\alpha \ \ and\ \ \angle DAC=\beta$$,

$$\angle ABD={{180}^{o}}-\left( \alpha+{{180}^{o}}-\theta \right)=\theta-\alpha =\beta$$,

From equation i/ii

$$\frac{BDsin\beta }{DCsin\alpha }=\frac{\sin \left( \theta +\beta \right)}{\sin \left( \theta -\alpha \right)}$$…..(iii)

We have $$\angle CAD={{180}^{0}}-\left( \theta +C \right)$$,

$$\angle ABC=B$$,

$$\angle ACD=C$$,

$$\angle BAD=\left( \theta -B \right)$$,

equation (iii) substitute above values

$$\frac{BD\sin B}{DC\sin C}=\frac{\sin \left( \theta -B \right)}{\sin \left( \theta +C \right)}$$,

BD/DC = m/n

$$\frac{m\sin B}{n\sin C}=\frac{\sin \left( \theta -B \right)}{\sin \left( \theta +C \right)}$$,

$$m\sin B\left( \sin \theta \cos C+\cos \theta \sin C \right)=n\sin C\left( \sin \theta \cos B-\cos \theta \sin B \right)$$,

dividing both sides by sinB sinC sinθ

m (cotC + cotθ) = n (cotB – cotC)

(m + n) cotθ = n cotB – m cotC