Problem 2

Let

I

be the incenter of a triangle

ABC

and let

\Gamma

be its circumcircle. Let the line

AI

intersect

\Gamma

again at

D

. Let

E

be a point on the arc

BDC

(the arc not containing

A

) and

F

a point on the side

BC

such that

\angle BAF=\angle CAE<\dfrac12\angle BAC

. Let

G

be the midpoint of the segment

IF

. Prove that the lines

DG

and

EI

intersect on

\Gamma

Step 6 of 6: Midline plus equal angles finish the concyclicity

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In plain words

$D$ and $G$ are the midpoints of $IA'$ and $IF$ , so $DG$ is a midline of triangle $IFA'$ parallel to $FA'$ . Sliding Step 5's equal angle along this parallel shows that the meeting point of $DG$ and $EI$ subtends equal angles to $D$ and $E$ , which is exactly the condition for lying on the circle through $A$ , $D$ , $E$ .

DG\parallel FA'\implies\angle GDA=\angle FA'A=\angle IEA

Detailed analysis

By Step 2, $D$ is the midpoint of $IA'$ , and by hypothesis $G$ is the midpoint of $IF$ ; hence in triangle $IFA'$ , segment $DG$ joins the midpoints of sides $IA'$ and $IF$ , so $DG\parallel FA'$ and $\angle GDA=\angle FA'A$ (corresponding angles cut by transversal $AA'=AD$ ). Combined with Step 5's $\angle FA'A=\angle IEA$ , this gives $\angle GDA=\angle IEA$ . Let $X$ be the intersection of lines $DG$ and $EI$ ; then $\angle XDA=\angle GDA=\angle IEA=\angle XEA$ , so $D$ and $E$ see segment $AX$ at equal angles, meaning $A$ , $D$ , $E$ , $X$ are concyclic. Since $A$ , $D$ , $E$ already determine $\Gamma$ , the point $X$ must lie on $\Gamma$ , proving that lines $DG$ and $EI$ meet on $\Gamma$ .