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Suppose $\triangle ABC$ and $\triangle A'B'C'$ are two similar but non congruent triangles such that $AB$ is parallel to $A'B'$, $AC$ is parallel to $A'C'$, and $BC$ is parallel to $B'C'$. Prove that lines $AA'$, $BB'$, and $CC'$ are concurrent.

I have proved the case where the point of intersection is past both triangles. I am trying to prove it when the point is in between the two triangles.

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Suppose the ratio of similarity is $$\frac{AB}{A'B'}=\frac{BC}{B'C'}=\frac{CA}{C'A'}=r$$. Let $X$ be the intersection of $AA'$ and $BB'$. Since $A'B'\parallel AB$, $\triangle XA'B'\sim XAB$ with scale factor $r$, and $\displaystyle\frac{XB}{XB'}=\frac{XA}{XA'}=r$. Similarly, if $Y$ is the intersection of $BB'$ and $CC'$, then we get $\frac{YB}{YB'}=r$. Since $X$ and $Y$ both lie on $BB'$, and $\frac{YB}{YB'}=\frac{XB}{XB'}$ (using directed lengths), then $X=Y$ and $AA'$,$BB'$,$CC'$ concur.

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This is Desargues' theorem: Corresponding sides of the two triangles are parallel, hence intersect in three points of $\ell_\infty$, the line at infinity. It follows that the three lines connecting two corresponding vertices intersect in a point $P$. The extra assumptions made here guarantee that $P\notin\ell_\infty$.

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