# Thread: Help with Feuerbach's circle proof using dilations

1. ## Help with Feuerbach's circle proof using dilations

I included the assignment as an attachment. I am trying to prove Feuerbach's Nine Point circle theorem using dilations. For the first question, I think that if ${\delta}(O)=O'$ if I let ${\delta}(Q)=Q'$ where Q and Q' are points on their respective circles, then I essentially have two triangles formed where side OQ has length r, radius of $C(O, r)$ and side O'Q' with length r', radius of $C(O',r'=kr)$. I am not sure if my reasoning is valid but if it is, how can I use this to deduce that Q' will always be on $C(O', r')$.

I really appreciate the help.

2. Originally Posted by ordinalhigh
I included the assignment as an attachment. I am trying to prove Feuerbach's Nine Point circle theorem using dilations. For the first question, I think that if ${\delta}(O)=O'$ if I let ${\delta}(Q)=Q'$ where Q and Q' are points on their respective circles, then I essentially have two triangles formed where side OQ has length r, radius of $C(O, r)$ and side O'Q' with length r', radius of $C(O',r'=kr)$. I am not sure if my reasoning is valid but if it is, how can I use this to deduce that Q' will always be on $C(O', r')$.

I really appreciate the help.
You have an arbitrary circle with center O and radius r, and an arbitrary point Q on that circle. And you have defined a circle with center O' (the image of O) and radius kr. So all you need to know is if Q' (the image of Q) is on the second circle. That will be true if the distance from O' to Q' is kr.
So you have defined two triangles XOQ and XO'Q'. Are they similar? If so, the ratio of the sides are all equal....

I think you can take it from there. I think if you draw diagrams for the other three problems you'll be able to do them, too.

Let me know if you have any more trouble.

3. Originally Posted by hollywood
You have an arbitrary circle with center O and radius r, and an arbitrary point Q on that circle. And you have defined a circle with center O' (the image of O) and radius kr. So all you need to know is if Q' (the image of Q) is on the second circle. That will be true if the distance from O' to Q' is kr.
So you have defined two triangles XOQ and XO'Q'. Are they similar? If so, the ratio of the sides are all equal....

I think you can take it from there. I think if you draw diagrams for the other three problems you'll be able to do them, too.

Let me know if you have any more trouble.
I think I understand it now. The similarity of the two triangles implies that the segment O'Q' is of length kr, so it must lie on the circle C(O',r').

I think that for the second one, I can use midline property to show that the line BC of a triangle ABC is a perpendicular bisector of the extended altitude from A to the point on the circumcircle since the segment HA' will be perpendicular to BC.

I think I have an idea of how the results will be used to prove the Nine point circle theorem but I am stuck on problem three. How can I show that the segments from the orthocenter to the diametral opposites of the points A, B, C, will bisect the sides of the triangle?

4. I finally have your result (too late, I guess).

Since AA' is a diameter and B is on the circle, ABA' is a right angle.
If you continue CH to a point X on AB, it is an altitude, so CX is perpendicular to AB.
So CH is parallel to A'B.

Since AA' is a diameter and C is on the circle, ACA' is a right angle.
If you continue BH to a point Y on AC, it is an altitude, so BY is perpendicular to AC.
So CA' is parallel to HB.

Thus the quadrilateral BHCA' is a parallelogram, and the diagonals of a parallelogram bisect each other, so HA' and BC bisect each other.

Are you taking geometry at a university? The assignment seems a little difficult for high school. What did you end up turning in? And what was your professor's solution?

- Hollywood