# Properties of Triangle – Theorem

## Properties of Triangle – 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 = m cotα – n cotβ

Proof:

BD/DC = m/n

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

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

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

From ΔABD,

From equation i/ii

$$\frac{BD/sin\alpha }{DC/sin\beta }=\frac{AD/sin\left( \theta \text{ }\text{ }\alpha \right)}{AD/sin\left( \theta \text{ }+\text{ }\beta \right)}$$,

$$\frac{BD/sin\beta }{DC/sin\alpha}=\frac{\sin \left( \theta +\beta \right)}{\sin \left( \theta -\alpha \right)}$$,

(since using sin(x+y) and sin(x-y) formulas)

$$\frac{BDsin\beta }{DCsin\alpha }=\frac{\sin \theta \cos \beta +\cos \theta \sin \beta }{\sin \theta \cos \alpha -\cos \theta \sin \alpha }$$,

BD/DC = m/n

$$\frac{m\ sin\beta }{n\ sin\alpha }=\frac{\sin \theta \cos \beta +\cos \theta \sin \beta }{\sin \theta \cos \alpha -\cos \theta \sin \alpha }$$,

m sin β(sin θ  cos α -cos θ sin α) = n sin α (sinθ cos β + cosθ sin β )

dividing both sides by sinα sinβ sinθ

m cot α – m cot θ = n cot β + n cot θ

m cot α – n cot β = n cot θ + m cot θ

cot θ (m+ n) = m cot α – n cot β

Example: If the median AD of triangle ABC makes an angle π/4 with side BC, then find the valve of

|cotB – cotC|.

solution:

by m- n theorem

cot θ (m+ n) = m cot α – n cot β

(BD+DC) cot π/4 = DC cotB – BD cotC

|cotB – cotC| = 2