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 4 of 6: Similar triangles from the equal angles at A

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In plain words

Since $E$ lies on the far arc, $\angle AEC=\angle ABC$ , and combined with the hypothesis $\angle BAF=\angle EAC$ this makes triangles $ABF$ and $AEC$ similar, so their sides multiply out to the same product as before.

AF\cdot AE=AB\cdot AC=AI\cdot AA'\implies\dfrac{AF}{AA'}=\dfrac{AI}{AE}

Detailed analysis

Points $A,B,E,C$ lie on $\Gamma$ with $B$ and $E$ on the same side of chord $AC$ (both on arc $BDC$ ), so the inscribed angle theorem gives $\angle AEC=\angle ABC=\angle ABF$ . Together with $\angle BAF=\angle EAC$ (from the problem's hypothesis, since $F$ lies on $BC$ ), triangles $ABF$ and $AEC$ are similar by AA, so $\dfrac{AB}{AE}=\dfrac{AF}{AC}$ , i.e. $AF\cdot AE=AB\cdot AC$ . Combining with Step 3's identity $AB\cdot AC=AI\cdot AA'$ gives $AF\cdot AE=AI\cdot AA'$ , which rearranges to $\dfrac{AF}{AA'}=\dfrac{AI}{AE}$ .