Transversals' Theorem

Famous theorems in 2D Geometry

Theorem. Whether triangle ABC and point D (BC), M (AB), N (AC) and {P} = MN AD. Then it has the relationship

Theorem transversal, geometry

Proof. We consider this figure:

theorem transversal, triangle

We have two cases:

First. MN || BC (which is simple to proof)

Second. MN neparalel BC

We build d || BC, A ∈ d and we consider {X} = MN ∩ BC and {Y} = d ∩ MN

Becouse d || BC we have, according Fundamental Theorem of Similarity,

triunghiuri asemenea

asemanare

asemanare

 And then we get:

 teorematransversalei

teorematransversalei

teorematransversalei

teorematransversalei

teorematransversalei

teorematransversalei

 teorematransversalei

 

Obs

If P is centre of gravity of the triangle we have

 

So though, the relation

becomes

frac{1}{2}cdot BC = frac{MB}{MA}cdot frac{BC}{2}+ frac{NC}{NA}cdot frac{BC}{2}

1=frac{MB}{MA}+frac{NC}{NA}

That means P - the centre of gravity of the triangle ABC, Transversals' Theorem becomes

1=frac{MB}{MA}+frac{NC}{NA}

Keywords: theorem, transversal, 2D geometry, proof

 

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