The statement of the theorem :A triangle ABC with the incenter I has its Schiffler point at the point of concurrence of the Euler lines of the four triangles BCI, CAI, ABI, and ABC. Schiffler's theorem states that these four lines all meet at a single point.
Question : I want to find a demonstration and process this result without using the idea of circle or perpendicularity. I have been trying for a few days to find a similar demonstration of this result, possibly using barycentric calculus. I only managed to discover some results that could take out the idea of bisectors and replaced the bisector with reports (Bisector Theorem) and a result that describes the line of Euler with the help of parallels. Specifically, I really want to prove this result using only concepts of parallelism and segment reports.
Help theorem : Let P be any point int he plane whatsoever. Let LALA be the line through the vertex AA that is parallel to the line through PP and the midpoint A′A′ of the opposite side BC. Let the lines LBLB and LCLC be constructed similarly. Then, the three line LALA, LBLB and LCLC have a point QQ in common, the points G,PG,P and QQ lie on the line, and GQ=2PGGQ=2PG. The more accurate question is: How can I prove the result of the 'theorem' Schiffler Point with the help of the theorem mentioned above or using other results, but only using the concept of parallelism and ratios (exactly in the affine space)?
Question : I want to find a demonstration and process this result without using the idea of circle or perpendicularity. I have been trying for a few days to find a similar demonstration of this result, possibly using barycentric calculus. I only managed to discover some results that could take out the idea of bisectors and replaced the bisector with reports (Bisector Theorem) and a result that describes the line of Euler with the help of parallels. Specifically, I really want to prove this result using only concepts of parallelism and segment reports.
Help theorem : Let P be any point int he plane whatsoever. Let LALA be the line through the vertex AA that is parallel to the line through PP and the midpoint A′A′ of the opposite side BC. Let the lines LBLB and LCLC be constructed similarly. Then, the three line LALA, LBLB and LCLC have a point QQ in common, the points G,PG,P and QQ lie on the line, and GQ=2PGGQ=2PG. The more accurate question is: How can I prove the result of the 'theorem' Schiffler Point with the help of the theorem mentioned above or using other results, but only using the concept of parallelism and ratios (exactly in the affine space)?