# 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

Proof. We consider this figure:

We have two cases:

First. (which is simple to proof)

Second.

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,

And then we get:

### Obs

If P is centre of gravity of the triangle we have

• D - mid of BC => BD = DC = BC / 2
• PD = 2 AD / 3 => PA = 2 • PD, i.e.:

So though, the relation

becomes

$\frac{1}{2}\cdot&space;BC&space;=&space;\frac{MB}{MA}\cdot&space;\frac{BC}{2}+&space;\frac{NC}{NA}\cdot&space;\frac{BC}{2}$

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

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

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

Keywords: theorem, transversal, 2D geometry, proof