Given a triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
, with
![I](/media/m/3/8/6/38689d6affa9ba35368ca4d3d76ea147.png)
as its incenter and
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
as its circumcircle,
![AI](/media/m/d/6/c/d6ca198d9456e8d2c2e0f932598e4a02.png)
intersects
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
again at
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
. Let
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
be a point on the arc
![BDC](/media/m/9/8/e/98ef4d3793b518113c2edb7d38c11e92.png)
, and
![F](/media/m/3/e/8/3e8bad5df716d332365fca76f53c1743.png)
a point on the segment
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
, such that
![\angle BAF=\angle CAE < \dfrac12\angle BAC](/media/m/e/0/0/e00e463fa5e8e4648ae659159a34002f.png)
. If
![G](/media/m/f/e/b/feb7f8fc95cee3c3a479382202e06a86.png)
is the midpoint of
![IF](/media/m/9/9/4/99490c823319f9160fbc22a10d37c9f8.png)
, prove that the meeting point of the lines
![EI](/media/m/3/3/6/336cf885dc5d1eecbca16a93299b3a4a.png)
and
![DG](/media/m/a/9/f/a9f3beec07bea1decbc805d9bbfda986.png)
lies on
![\Gamma](/media/m/4/e/0/4e08987e1d0700578a2eb5c2fc65dc3b.png)
.
Proposed by Tai Wai Ming and Wang Chongli, Hong Kong
%V0
Given a triangle $ABC$, with $I$ as its incenter and $\Gamma$ as its circumcircle, $AI$ intersects $\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$, such that $\angle BAF=\angle CAE < \dfrac12\angle BAC$. If $G$ is the midpoint of $IF$, prove that the meeting point of the lines $EI$ and $DG$ lies on $\Gamma$.
Proposed by Tai Wai Ming and Wang Chongli, Hong Kong