(CGMO 2012) The incircle of a triangle ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of triangle BCI where \(I\) denotes the incenter of \(\bigtriangleup ABC\). Prove that \(\angle ODB = \angle OEC \)
We will first prove this lemma.
Let be a triangle with incenter , A-excenter , and denote the midpoint of arc to be , then points are concyclic with as the circle center.
(We will denote angle , , , as and , respectively.)
Also note that
Proof: First notice that is the intersection of the angle bisector of and perpendicular bisector of .
First we prove that .
Since the perpendicular of to is the perpendicular bisector of , we know that .
Because are angle bisectors, we know that and . By Exterior Angle Theorem, we know that . If we prove that we have proved the segments are equal.
Notice that because they substend the same arc. Therefore this implies that . Thus, by isosceles triangles we get that . This tells us that are concyclic with as a center.
Now, we must prove that also lies on this circle.
Recall that the is the intersection of the exterior-angle bisectors of and . Therefore if you extend and and construct the angle bisector of the exterior angle, it intersects at .
Denote as a point on ray past B.
By Exterior Angle Theorem, we have that .
Since is the angle bisector, we have that . Since , this implies that .
We proved that .
Again by exterior angle theorem, since , we have that which implies that is isoscles which implies that . Therefore must be concyclic with because it's radius to the center, , is the same. Therefore, we have proved the lemma.
Okay good. Now let's have fun.
Restatement: The incircle of a triangle ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of triangle BCF where denotes the incenter of . Prove that
By the lemma, we know that the circumcenter, , of is actually the midpoint of arc BC of the circumcircle of .
To prove , it suffices to prove that .
Since is on the angle bisector, we know that . We also know that because they are both tangents to the incircle. Additionally, and share side . Therefore, by , we have that .
This implies that and we are done.