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 2 of 6: The excenter lemma: D is equidistant from B, I, C, and the excenter

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In plain words

The classical incenter–excenter lemma says the point where the internal bisector from $A$ meets $\Gamma$ is equally far from $B$ , $I$ , $C$ , and the $A$ -excenter $A'$ , because $BI\perp BA'$ makes $IA'$ a diameter of the circle centered at that point.

DI=DB=DC=DA'

Detailed analysis

Let $A'$ be the $A$ -excenter. Angle chasing gives $\angle DBI=\angle DIB=\tfrac12\angle A+\tfrac12\angle B$ , so $DB=DI$ , and symmetrically $DC=DI$ . Since $BI$ and $BA'$ are the internal and external bisectors of angle $B$ , they satisfy $BI\perp BA'$ , so $IA'$ is a diameter of a circle centered on line $AI$ ; that center must be $D$ itself, giving $DI=DA'$ . Hence $D$ is the midpoint of $IA'$ , and $DI=DB=DC=DA'$ .