if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?
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D is the circumcenter of BIC
If triangle QRS has incenter, what is the incenter of triangle xyz
circleseuclidean-geometrygeometrytriangles
Related Solutions
Here's a nice proof by contradiction.
Let the incenter $I$ lie on the Euler line of $ABC$.
It is known that orthocenter $H$ and circumcenter $O$ are isogonal conjugates, i.e. $AI$ is the bisector of angle $HAO$.
So, (if the point $A$ does not lie on Euler line) $HA/AO=HI/IO$ (angle bisector theorem). Also $HB/BO=HC/CO=HI/IO=HA/AO$. And we know that all points $X$, such that $YX/ZX=const$, lie on a circle with center on the line $YZ$ (Appolonius circle)
So, $A, B, C$ and $I$ lies on the same circle, and that cannot be true. We have assumed that all of points $A, B, C$ don't lie on the Euler line, so, one of them lies on Euler line and that means $ABC$ is isosceles.
Here is a possible proof "without inversion". (Note: Generally i am against settings of problems, which discriminate some structural part of mathematics, and ask for a solution "without" some ingredient, which would make the solution straightforward, simple, and easy to remember. But in this case i will accept the challenge. There will be some similitude argument instead... well, same two circles as in the solution using inversion are the main actors.)
Note that even the main wiki page for the inversion uses this problem to illustrate the use and usefulness of the inversion:
https://en.wikipedia.org/wiki/Inversive_geometry#Application
I will slightly change the notation to fit the order in the alphabet. So $M$ is not on $AB$, but on the side opposite to the vertex $A$, and similarly for $N,P$. (Else i would commit errors.) So let us state explicitly...
Let $\Delta ABC$ be a triangle. Let $I$ be the incenter, the intersection of the bisectors of the angles of the triangle. The incircle $(I)$ touches the sides of $\Delta ABC$ in the points $M\in BC$, $N\in CA$, and $P\in AB$.
Let $S,T,U$ be the intersections of the the angle bisectors $AI$, $BI$, $CI$ with the segments respectively perpendicular on them, $NP$, $PM$, $MN$.
Let $O$ be the circumcenter of $\Delta ABC$, the center of the circle $(ABC)$.
Let $o$ be the circumcenter of $\Delta MNP$, the center of the circle $(MNP)$. Of course, $o=I$, but it may be useful to use the notation $o$ when addressing this point for its quality of being the center of $(MNP)$. We use in the notation a small letter as possible, and with the same convention let $h$, $g$, $9$ be respectively for $\Delta MNP$ the orthocenter, the centroid, and the center of the nine point (Euler) circle $(STU)$. Here, $g=MS\cap NT\cap PU$, and $9$ is the mid point of $oh$.
We denote by $A'$ the intersection of the bisector $AI$ with the circle $(O)=(ABC)$. Construct $B',C'$ similarly.
Let $M'$ be the mid point of the segment $Mh$, so $M'\in(STU)$.
The we have:
(1) The Euler line (e) of $\Delta MNP$ is passing through $o=I$, $9$, $g$, $h$, in particular these points are colinear.
(2) $OA'\| oM\|S9M'$, and similarly $OB'\| oN\|T9N'$, $OC'\| oP\|U9P'$.
(3) The triangles $\Delta STU$ and $\Delta A'B'C'$ are perspective / similar / homothetic, the center of the perspectivity being $o=I$, the intersection of the bisectors $ASIA'$, $BTIB'$, $CUIC'$.
(4) The circles $(STU)$ and $(A'B'C')=(ABC)$ respect the same homothety. In particular, the center of homothety $o$ is on the line their centers $9$ and $O$.
(5) The points $o=I$, $9$, $g$, $h$, $O$ are on the Euler line $(e)$ of $\Delta MNP$.
Proof: All points are clear. (2) deserves maybe a slight explanation. The lines $oM=IM$, and $OA'$ are perpendicular on $BC$. ($M$ is the tangency point of $(I)$, and $OA'$ is the side bisector of $BC$.) So $oM\|OA'$. Then in $\Delta MNP$ we have the parallelogram $oMM'S$. (For instance, $IS$, $MM'$ are perpendicular on $NP$, it remains to show they have the same length. Use now that $9$ is the mid point of $oh$, $9$ is also the mid point of the diameter $M'S$, so $ShM'o$ parallelogram, so $SI=So=hM'=M'M$.)
$\square$
Best Answer
Join $AI$ and extend till it intersects circumcircle. If possible Ray$AI$ does not intersect $D$ but at $D'$ . Join $D'B$ and $D'C$ . By angle chasing, $\angle BID' = \angle IBD'$ And $\angle CID' = \angle ICD'$. Therefore, $BD'$ = $ID'$ = $CD'$. $\implies$ $D'$ is circumcentre of $\triangle BIC$ . As given $D$ is circumcentre of $\triangle BIC$ . Contradiction to considerations, $D$ coincides $D'$. Hence, proved.