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 3 of 6: A product identity linking the incenter and excenter distances

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In plain words

Both $AI$ and $AA'$ are half-angle distances from $A$ , and combined with the law of sines for $AB$ and $AC$ their product collapses to the same expression, giving the stated identity.

AB\cdot AC=AI\cdot AA'

Detailed analysis

By the law of sines, $AB=2R\sin C$ and $AC=2R\sin B$ , where $R$ is the circumradius. Two standard incenter/excenter distance formulas give $AI=4R\sin\tfrac{B}{2}\sin\tfrac{C}{2}$ and $AA'=4R\cos\tfrac{B}{2}\cos\tfrac{C}{2}$ . Multiplying, $AI\cdot AA'=16R^2\sin\tfrac{B}{2}\sin\tfrac{C}{2}\cos\tfrac{B}{2}\cos\tfrac{C}{2}=4R^2\sin B\sin C=AB\cdot AC$ , using $\sin\theta=2\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$ .