I have a triangle $ABC$ with incenter $I$. $AI$ extended meets the circumcircle of $ABC$ at $M$. Prove that $CM=BM=IM$.
I was able to prove that $CM=BM$ taking advantage of the fact that the perpendicular bisector of $BC$ also meets at $M$, but I don't see how to get the $IM$ part. Can anyone help? :)
On
I think you could work this out using vectors with the following steps:
1) Determine the line L intersecting A and I.
2) Use the midpoints of the triangle sides to determine the circumcenter S (See 'http://www.mathopenref.com/trianglecircumcenter.html').
3) Determine the radius R of the circumcenter circle C to produce a formula for C.
4) Determine the intersection of circle C with line L. One point will be the triangle vertex A while the other will be point M (See 'http://mathworld.wolfram.com/Circle-LineIntersection.html').
5) Determine the distance between I and M using a distance function d. You are likely just going to use d((x1,y1),(x2,y2)) = sqrt((x1-x2)^2+(y1-y2)^2) .
6) Verify whether or not d(I,M) == d(B,M) == d(C,I) .
$$ C \hat AI =B \hat AM = B \hat CM \\ M \hat IC = I \hat AC + I \hat CA \\ M \hat CI = B \hat AM + I \hat CB $$ since we also have $$ I \hat CA = I \hat CB $$ the triangle $MCI$ is isosceles