Let O be circumcenter of a triangle ABC

[Math Tutor]

Let O be the circumcenter of a triangle ABC.
A line through O intersects BC at M and CA at N.
R and S are midpoints of AM and BN.
Prove that [tex]\angle{ROS}=\angle{ACB}[/tex]

[Math Tutor]

Author: ins-

[Guest]

The circumcentre O is the centre of the circumscribed circle...the one outside passing through point A, B, and C.
O is the point of intersection of the perpendicular bisectors of 2 or the 3 sides.
AM is a line through O from A and intersects with BC somewhere.
BN is a line through O from B and intersects AC somewhere.
R is mid point of the line AM and is on the line AM
S is the midpoint of BN and is on the line BN
AM and BN both pass through O.
So ROS is same angle as AOB which is the line at the centre of the circumscribed circle subtended from the arc/chord AB.
ACB is the angle at the circumference suntended from the arc/chord AB.
So ROS is twice ACB.....and not equal as stated in question...????...

I am sure someone will have an answer.................??

............................Simple............................

[Guest]

Correction to last post...sorry............

So ROS is same angle as AOB which is the ANGLE at the centre of the circumscribed circle subtended from the arc/chord AB.
ACB is the angle at the circumference SUBTENDED from the arc/chord AB.
So ROS is twice ACB.....and not equal as stated in question...????...

I am sure someone will have an answer.................??

............................Simple............................

[Guest]

Diagram (and solution) can be found in the link.
http://www.artofproblemsolving.com/blog/97582

R. Baber.

[Guest]

Yes, got it, It's a different question........

[Guest]

Please note, I don't pretend to be creator of the problem I found it interesting and it was the reason I proposed it. /ins-/