Thales properties with similar triangles

Question

Let :

• $\Delta ABC$ is a triangle
• $0<\beta<1$
• We put $M$ and $N$ two points such that $AM=\beta AB$ and $AN=\beta AC$

We want to prove that $(MN)$ parallel to $(BC)$ and $MN= \beta BC$. I made a proof using only similar triangles. Is the proof correct ?

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• We draw the unique line parallel to $(AB)$ and lying on $C$. Let $E$ be one point of this line. $E \ne C$

• $D$ is the intersection of $(MN)$ and $(CE)$

• $\Delta AMN$ and $\Delta NDC$ are similar. $\widehat{MAN}=\widehat{NDC}$ and $\widehat{AMN}=\widehat{NDC}$ therefore :

• $ \exists \, 0<\alpha<1 $ such that $AN = \alpha AC$ and $MN= \alpha MD$

• so $\alpha =\beta$ and $MD= BC$

• $\Delta AMN$ and $\Delta ABC$ are similar because their sides are proportionnal

• $\widehat{AMN}= \widehat{ABC}$

• Finally $(MN)$ is parallel to $(BC)$

Answer 1

You should correct $=$ to $=M$, first of all. But then, I don't understand how you got $MD=BC$.

Apart from such details, I have a major objection to your reasoning: you are using SSS similarity criterion (all the corresponding sides are in the same ratio), which is usually proved by the aid of this "Thales property" you are trying to prove. (And, by the way, you could prove immediately the similarity between $AMN$ and $ABC$ by SAS criterion, against which the same objection can be raised.)

To avoid circularity, you should then exhibit a proof of that criterion which doesn't make use of Thales property.

Answer 2

If two lines are intersected by a third transversal line and the alternate angles are equal ( by triangle similarity here ), then the two lines are parallel.. due to Euclid Proposition 27.