Subfield: Geometry


Question: A point $D$ is chosen inside an acute-angled triangle $A B C$ with $A B>A C$ so that $\angle B A D=\angle D A C$. A point $E$ is constructed on the segment $A C$ so that $\angle A D E=\angle D C B$. Similarly, a point $F$ is constructed on the segment $A B$ so that $\angle A D F=\angle D B C$. A point $X$ is chosen on the line $A C$ so that $C X=B X$. Let $O_{1}$ and $O_{2}$ be the circumcentres of the triangles $A D C$ and $D X E$. Prove that the lines $B C, E F$, and $O_{1} O_{2}$ are concurrent.

Solution:
Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\angle B A D=\angle D A C$, the point $Q$ lies on $A D$. Then $\angle Q B A=\angle D B C=\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \cdot A B=A D \cdot A Q=A E \cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.

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Let $T$ be the intersection of $B C$ and $F E$.

Claim. $T D^{2}=T B \cdot T C=T F \cdot T E$.

Proof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get

$$
\begin{aligned}
& \angle B D F=\angle A F D-\angle A B D=\left(180^{\circ}-\angle F A D-\angle F D A\right)-(\angle A B C-\angle D B C) \\
= & 180^{\circ}-\angle F A D-\angle A B C=180^{\circ}-\angle D A E-\angle F E A=\angle F E D+\angle A D E=\angle F E D+\angle D C B,
\end{aligned}
$$

which implies the desired tangency.

Since the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \cdot T F=T B \cdot T C$.



Let $T A$ intersect the circle $(A B C)$ again at $M$. Due to the circles $(B C E F)$ and $(A M C B)$, and using the above Claim, we get $T M \cdot T A=T F \cdot T E=T B \cdot T C=T D^{2}$; in particular, the points $A, M, E$, and $F$ are concyclic.

Under the inversion with centre $T$ and radius $T D$, the point $M$ maps to $A$, and $B$ maps to $C$, which implies that the circle $(M B D)$ maps to the circle $(A D C)$. Their common point $D$ lies on the circle of the inversion, so the second intersection point $K$ also lies on that circle, which means $T K=T D$. It follows that the point $T$ and the centres of the circles $(K D E)$ and $(A D C)$ lie on the perpendicular bisector of $K D$.

Since the center of $(A D C)$ is $O_{1}$, it suffices to show now that the points $D, K, E$, and $X$ are concyclic (the center of the corresponding circle will be $O_{2}$ ).

The lines $B M, D K$, and $A C$ are the pairwise radical axes of the circles $(A B C M),(A C D K)$ and $(B M D K)$, so they are concurrent at some point $P$. Also, $M$ lies on the circle $(A E F)$, thus

$$
\begin{aligned}
\sphericalangle(E X, X B) & =\sphericalangle(C X, X B)=\sphericalangle(X C, B C)+\sphericalangle(B C, B X)=2 \sphericalangle(A C, C B) \\
& =\sphericalangle(A C, C B)+\sphericalangle(E F, F A)=\sphericalangle(A M, B M)+\sphericalangle(E M, M A)=\sphericalangle(E M, B M),
\end{aligned}
$$

so the points $M, E, X$, and $B$ are concyclic. Therefore, $P E \cdot P X=P M \cdot P B=P K \cdot P D$, so the points $E, K, D$, and $X$ are concyclic, as desired.



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Common remarks. Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\angle B A D=\angle D A C$, the point $Q$ lies on $A D$. Then $\angle Q B A=\angle D B C=\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \cdot A B=A D \cdot A Q=A E \cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.

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Let $T$ be the intersection of $B C$ and $F E$.

Claim. $T D^{2}=T B \cdot T C=T F \cdot T E$.

Proof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get

$$
\begin{aligned}
& \angle B D F=\angle A F D-\angle A B D=\left(180^{\circ}-\angle F A D-\angle F D A\right)-(\angle A B C-\angle D B C) \\
= & 180^{\circ}-\angle F A D-\angle A B C=180^{\circ}-\angle D A E-\angle F E A=\angle F E D+\angle A D E=\angle F E D+\angle D C B,
\end{aligned}
$$

which implies the desired tangency.

Since the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \cdot T F=T B \cdot T C$.




We use only the first part of the Common remarks, namely, the facts that the tuples $(C, D, Q, E)$ and $(B, C, E, F)$ are both concyclic. We also introduce the point $T=$ $B C \cap E F$. Let the circle $(C D E)$ meet $B C$ again at $E_{1}$. Since $\angle E_{1} C Q=\angle D C E$, the arcs $D E$ and $Q E_{1}$ of the circle $(C D Q)$ are equal, so $D Q \| E E_{1}$.

Since $B F E C$ is cyclic, the line $A D$ forms equal angles with $B C$ and $E F$, hence so does $E E_{1}$. Therefore, the triangle $E E_{1} T$ is isosceles, $T E=T E_{1}$, and $T$ lies on the common perpendicular bisector of $E E_{1}$ and $D Q$.

Let $U$ and $V$ be the centres of circles $(A D E)$ and $(C D Q E)$, respectively. Then $U O_{1}$ is the perpendicular bisector of $A D$. Moreover, the points $U, V$, and $O_{2}$ belong to the perpendicular bisector of $D E$. Since $U O_{1} \| V T$, in order to show that $O_{1} O_{2}$ passes through $T$, it suffices to show that

$$
\frac{O_{2} U}{O_{2} V}=\frac{O_{1} U}{T V}
\tag{1}
$$

Denote angles $A, B$, and $C$ of the triangle $A B C$ by $\alpha, \beta$, and $\gamma$, respectively. Projecting onto $A C$ we obtain

$$
\frac{O_{2} U}{O_{2} V}=\frac{(X E-A E) / 2}{(X E+E C) / 2}=\frac{A X}{C X}=\frac{A X}{B X}=\frac{\sin (\gamma-\beta)}{\sin \alpha}
\tag{2}
$$

The projection of $O_{1} U$ onto $A C$ is $(A C-A E) / 2=C E / 2$; the angle between $O_{1} U$ and $A C$ is $90^{\circ}-\alpha / 2$, so

$$
\frac{O_{1} U}{E C}=\frac{1}{2 \sin (\alpha / 2)}
\tag{3}
$$

Next, we claim that $E, V, C$, and $T$ are concyclic. Indeed, the point $V$ lies on the perpendicular bisector of $C E$, as well as on the internal angle bisector of $\angle C T F$. Therefore, $V$ coincides with the midpoint of the arc $C E$ of the circle $(T C E)$.

Now we have $\angle E V C=2 \angle E E_{1} C=180^{\circ}-(\gamma-\beta)$ and $\angle V E T=\angle V E_{1} T=90^{\circ}-\angle E_{1} E C=$ $90^{\circ}-\alpha / 2$. Therefore,

$$
\frac{E C}{T V}=\frac{\sin \angle E T C}{\sin \angle V E T}=\frac{\sin (\gamma-\beta)}{\cos (\alpha / 2)}
\tag{4}
$$

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Recalling (2) and multiplying (3) and (4) we establish (1):

$$
\frac{O_{2} U}{O_{2} V}=\frac{\sin (\gamma-\beta)}{\sin \alpha}=\frac{1}{2 \sin (\alpha / 2)} \cdot \frac{\sin (\gamma-\beta)}{\cos (\alpha / 2)}=\frac{O_{1} U}{E C} \cdot \frac{E C}{T V}=\frac{O_{1} U}{T V}
$$
Let $Q$ be the isogonal conjugate of $D$ with respect to the triangle $A B C$. Since $\angle B A D=\angle D A C$, the point $Q$ lies on $A D$. Then $\angle Q B A=\angle D B C=\angle F D A$, so the points $Q, D, F$, and $B$ are concyclic. Analogously, the points $Q, D, E$, and $C$ are concyclic. Thus $A F \cdot A B=A D \cdot A Q=A E \cdot A C$ and so the points $B, F, E$, and $C$ are also concyclic.

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Let $T$ be the intersection of $B C$ and $F E$.

Claim. $T D^{2}=T B \cdot T C=T F \cdot T E$.

Proof. We will prove that the circles $(D E F)$ and $(B D C)$ are tangent to each other. Indeed, using the above arguments, we get

$$
\begin{aligned}
& \angle B D F=\angle A F D-\angle A B D=\left(180^{\circ}-\angle F A D-\angle F D A\right)-(\angle A B C-\angle D B C) \\
= & 180^{\circ}-\angle F A D-\angle A B C=180^{\circ}-\angle D A E-\angle F E A=\angle F E D+\angle A D E=\angle F E D+\angle D C B,
\end{aligned}
$$

which implies the desired tangency.

Since the points $B, C, E$, and $F$ are concyclic, the powers of the point $T$ with respect to the circles $(B D C)$ and $(E D F)$ are equal. So their radical axis, which coincides with the common tangent at $D$, passes through $T$, and hence $T D^{2}=T E \cdot T F=T B \cdot T C$.


Notice that $\angle A Q E=\angle Q C B$ and $\angle A Q F=\angle Q B C$; so, if we replace the point $D$ with $Q$ in the problem set up, the points $E, F$, and $T$ remain the same. So, by the Claim, we have $T Q^{2}=T B \cdot T C=T D^{2}$.

Thus, there exists a circle $\Gamma$ centred at $T$ and passing through $D$ and $Q$. We denote the second meeting point of the circles $\Gamma$ and $(A D C)$ by $K$. Let the line $A C$ meet the circle $(D E K)$ again at $Y$; we intend to prove that $Y=X$. this will yield that the point $T$, as well as the centres $O_{1}$ and $O_{2}$, all lie on the perpendicular bisector of $D K$.

Let $L=A D \cap B C$. We perform an inversion centred at $C$; the images of the points will be denoted by primes, e.g., $A^{\prime}$ is the image of $A$. We obtain the following configuration, constructed in a triangle $A^{\prime} C L^{\prime}$.

The points $D^{\prime}$ and $Q^{\prime}$ are chosen on the circumcircle $\Omega$ of $A^{\prime} L^{\prime} C$ such that $\sphericalangle\left(L^{\prime} C, D^{\prime} C\right)=$ $\sphericalangle\left(Q^{\prime} C, A^{\prime} C\right)$, which means that $A^{\prime} L^{\prime} \| D^{\prime} Q^{\prime}$. The lines $D^{\prime} Q^{\prime}$ and $A^{\prime} C$ meet at $E^{\prime}$.

A circle $\Gamma^{\prime}$ centred on $C L^{\prime}$ passes through $D^{\prime}$ and $Q^{\prime}$. Notice here that $B^{\prime}$ lies on the segment $C L^{\prime}$, and that $\angle A^{\prime} B^{\prime} C=\angle B A C=2 \angle L A C=2 \angle A^{\prime} L^{\prime} C$, so that $B^{\prime} L^{\prime}=B^{\prime} A^{\prime}$, and $B^{\prime}$ lies on the perpendicular bisector of $A^{\prime} L^{\prime}$ (which coincides with that of $D^{\prime} Q^{\prime}$ ). All this means that $B^{\prime}$ is the centre of $\Gamma^{\prime}$.

Finally, $K^{\prime}$ is the second meeting point of $A^{\prime} D^{\prime}$ and $\Gamma^{\prime}$, and $Y^{\prime}$ is the second meeting point of the circle $\left(D^{\prime} K^{\prime} E^{\prime}\right)$ and the line $A^{\prime} E^{\prime}$, We have $\sphericalangle\left(Y^{\prime} K^{\prime}, K^{\prime} A^{\prime}\right)=\sphericalangle\left(Y^{\prime} E^{\prime}, E^{\prime} D^{\prime}\right)=$ $\sphericalangle\left(Y^{\prime} A^{\prime}, A^{\prime} L^{\prime}\right)$, so $A^{\prime} L^{\prime}$ is tangent to the circumcircle $\omega$ of the triangle $Y^{\prime} A^{\prime} K^{\prime}$.

Let $O$ and $O^{*}$ be the centres of $\Omega$ and $\omega$, respectively. Then $O^{*} A^{\prime} \perp A^{\prime} L^{\prime} \perp B^{\prime} O$. The projections of vectors $\overrightarrow{O^{*} A^{\prime}}$ and $\overrightarrow{B^{\prime} O}$ onto $K^{\prime} D^{\prime}$ are equal to $\overrightarrow{K^{\prime} A^{\prime}} / 2=\overrightarrow{K^{\prime} D^{\prime}} / 2-\overrightarrow{A^{\prime} D^{\prime}} / 2$. So $\overrightarrow{O^{*} A^{\prime}}=\overrightarrow{B^{\prime} O}$, or equivalently $\overrightarrow{A^{\prime} O}=\overrightarrow{O^{*} B^{\prime}}$. Projecting this equality onto $A^{\prime} C$, we see that the projection of $\overrightarrow{O^{*} B^{\prime}}$ equals $\overrightarrow{A^{\prime} C} / 2$. Since $O^{*}$ is projected to the midpoint of $A^{\prime} Y^{\prime}$, this yields that $B^{\prime}$ is projected to the midpoint of $C Y^{\prime}$, i.e., $B^{\prime} Y^{\prime}=B^{\prime} C$ and $\angle B^{\prime} Y^{\prime} C=\angle B^{\prime} C Y^{\prime}$. In the original figure, this rewrites as $\angle C B Y=\angle B C Y$, so $Y$ lies on the perpendicular bisector of $B C$, as desired.

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