Subfield: Geometry


Question: Let $I$ be the incenter of a triangle $A B C$ and $\Gamma$ be its circumcircle. Let the line $A I$ intersect $\Gamma$ at a point $D \neq A$. Let $F$ and $E$ be points on side $B C$ and $\operatorname{arc} B D C$ respectively such that $\angle B A F=\angle C A E<\frac{1}{2} \angle B A C$. Finally, let $G$ be the midpoint of the segment $I F$. Prove that the lines $D G$ and $E I$ intersect on $\Gamma$.

Solution:
Let $X$ be the second point of intersection of line $E I$ with $\Gamma$, and $L$ be the foot of the bisector of angle $B A C$. Let $G^{\prime}$ and $T$ be the points of intersection of segment $D X$ with lines $I F$ and $A F$, respectively. We are to prove that $G=G^{\prime}$, or $I G^{\prime}=G^{\prime} F$. By the Menelaus theorem applied to triangle $A I F$ and line $D X$, it means that we need the relation

$$
1=\frac{G^{\prime} F}{I G^{\prime}}=\frac{T F}{A T} \cdot \frac{A D}{I D}, \quad \text { or } \quad \frac{T F}{A T}=\frac{I D}{A D}
$$

Let the line $A F$ intersect $\Gamma$ at point $K \neq A$ (see Fig. 1); since $\angle B A K=\angle C A E$ we have $\overparen{B K}=\overparen{C E}$, hence $K E \| B C$. Notice that $\angle I A T=\angle D A K=\angle E A D=\angle E X D=\angle I X T$, so the points $I, A, X, T$ are concyclic. Hence we have $\angle I T A=\angle I X A=\angle E X A=\angle E K A$, so $I T\|K E\| B C$. Therefore we obtain $\frac{T F}{A T}=\frac{I L}{A I}$.

Since $C I$ is the bisector of $\angle A C L$, we get $\frac{I L}{A I}=\frac{C L}{A C}$. Furthermore, $\angle D C L=\angle D C B=$ $\angle D A B=\angle C A D=\frac{1}{2} \angle B A C$, hence the triangles $D C L$ and $D A C$ are similar; therefore we get $\frac{C L}{A C}=\frac{D C}{A D}$. Finally, it is known that the midpoint $D$ of $\operatorname{arc} B C$ is equidistant from points $I$, $B, C$, hence $\frac{D C}{A D}=\frac{I D}{A D}$.

Summarizing all these equalities, we get

$$
\frac{T F}{A T}=\frac{I L}{A I}=\frac{C L}{A C}=\frac{D C}{A D}=\frac{I D}{A D}
$$

as desired.

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Fig. 1

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Fig. 2
As in the previous solution, we introduce the points $X, T$ and $K$ and note that it suffice to prove the equality

$$
\frac{T F}{A T}=\frac{D I}{A D} \quad \Longleftrightarrow \quad \frac{T F+A T}{A T}=\frac{D I+A D}{A D} \quad \Longleftrightarrow \quad \frac{A T}{A D}=\frac{A F}{D I+A D} .
$$

Since $\angle F A D=\angle E A I$ and $\angle T D A=\angle X D A=\angle X E A=\angle I E A$, we get that the triangles $A T D$ and $A I E$ are similar, therefore $\frac{A T}{A D}=\frac{A I}{A E}$.

Next, we also use the relation $D B=D C=D I$. Let $J$ be the point on the extension of segment $A D$ over point $D$ such that $D J=D I=D C$ (see Fig. 2). Then $\angle D J C=$ $\angle J C D=\frac{1}{2}(\pi-\angle J D C)=\frac{1}{2} \angle A D C=\frac{1}{2} \angle A B C=\angle A B I$. Moreover, $\angle B A I=\angle J A C$, hence triangles $A B I$ and $A J C$ are similar, so $\frac{A B}{A J}=\frac{A I}{A C}$, or $A B \cdot A C=A J \cdot A I=(D I+A D) \cdot A I$.

On the other hand, we get $\angle A B F=\angle A B C=\angle A E C$ and $\angle B A F=\angle C A E$, so triangles $A B F$ and $A E C$ are also similar, which implies $\frac{A F}{A C}=\frac{A B}{A E}$, or $A B \cdot A C=A F \cdot A E$.

Summarizing we get

$$
(D I+A D) \cdot A I=A B \cdot A C=A F \cdot A E \quad \Rightarrow \quad \frac{A I}{A E}=\frac{A F}{A D+D I} \quad \Rightarrow \quad \frac{A T}{A D}=\frac{A F}{A D+D I}
$$

as desired.