How can we prove that A', B', and C' are collinear in triangle ABC?

[SonyDvDPro]

1 ) Let $$P$$ be an arbitary point in the plane of triangle $$\vartriangle ABC$$ , let $$A'$$ be a point on $$BC$$ such that $$A'B \bot PA$$
, Define $$B' , C'$$ in the same way,P rove that $$A' , B' ,C'$$ are collinear.

 

[dodo]

I'm not sure the statement of the problem is correct; if A' is a point on segment BC, then A'B is a subsegment of BC, and the choice of A' does not affect at all the (possible or not) parallel condition with respect to PA, since all P, A, B and C are fixed beforehand.

 

[tacman]

Nice problem! To prove that A', B', and C' are collinear, we can use the concept of perpendicular bisectors. Since A'B is perpendicular to PA, it must also be the perpendicular bisector of PA. Similarly, B'C and C'A are also perpendicular bisectors of PB and PC respectively.

Now, since the perpendicular bisectors of a triangle intersect at a single point (known as the circumcenter), we can conclude that A', B', and C' all intersect at the circumcenter of triangle ABC.

Since all three points lie on the same line (the line passing through the circumcenter), we can say that A', B', and C' are collinear. This proves the desired result.

Another way to prove this would be to use the concept of similarity. Since A'B is perpendicular to PA, we can say that triangle A'PB is similar to triangle A'PA. Similarly, triangle B'PC is similar to triangle B'PB and triangle C'PA is similar to triangle C'PC.

Since all three triangles have a common angle (angle P), we can say that they are all similar. This means that the ratio of their corresponding sides will be equal.

Therefore, we can say that A'B/PA = B'C/PB = C'A/PC.

Now, using the transitive property, we can say that A'B/PA = B'C/PB = C'A/PC = A'B'/PB' = B'C'/PC' = C'A'/PA'.

This shows that A', B', and C' are collinear, since they all have the same ratio to their corresponding sides.

Overall, there are multiple ways to prove that A', B', and C' are collinear. The concept of perpendicular bisectors and similarity are just two examples. This is a great problem that helps us understand the properties of triangles and their corresponding points.