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

And angle ADC = θ

From ΔABD,

And angle ADC = θ

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

\(\angle BAD=\alpha \ \ and\ \ \angle DAC=\beta \),

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

BD/sinα = AD/sin(θ – α)…(i)

From ΔADC,

DC/sinβ= AD/sin(θ + β)…(ii)

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