51st Ukrainian National Mathematical Olympiad, 4th Round

We will prove that $BX$ is a "simedian" of $△ABC$ (a cevian symmetric to the median with respect to the bisector, all drawn from the vertex $B$). By a known property of the "simedian" (which we will also prove) $\frac{CX}{AX}=\frac{C{B}^2}{A{B}^2}$. By the angle bisector theorem for $FX:\frac{CX}{AX}=\frac{FC}{FA}$ (fig. 31).

$△AFC$ is similar to $△BFA$ since they have a common angle $\angle CFA$, and $\angle CAF=\angle FBA$. So, $\frac{FC}{AF}=\frac{CX}{AX}=\frac{C{B}^2}{A{B}^2}$. $$\frac{AC}{AB}=\frac{FC}{AF}=\frac{CX}{AX}=\frac{C{B}^2}{A{B}^2},$$ which provides the required $$AB\cdot AC=C{B}^2.$$

Now we will prove the necessary properties of the “simedian”. 1) Let $AS=s$ be the “simedian”, as on fig. 32., $AL=l$ be the bisector, $AM=m$ — the median of $△ABC$, then $\angle MAL=\angle LAS$. Consider the quotient of the areas of the triangles $ASC$ and $AMB$ : $$\frac{S_{AMB}}{S_{ASC}}=\frac{\frac{1}{2}mc\sin \angle BAM}{\frac{1}{2}sb\sin \angle SAC}=\frac{cm}{bs},$$ because the angles $\angle BAM$ and $\angle SAC$ are equal (fig. 32). Similarly, consider the areas of the triangles $ASB$ and $AMC$ to obtain that $$\frac{S_{AMC}}{S_{ABS}}=\frac{bm}{cs}.$$ Since $AM$ is a median, we have that $S_{AMB}=S_{AMC}$, and so dividing the equalities we get: $$\frac{S_{ABS}}{S_{ASC}}=\frac{c^2}{b^2}=\frac{BS}{SC},$$ as required.

Let $O$ be the center of the circle. Then in the right triangle $FOC$ we have $OM\cdot OF=O{C}^2=R^2=O{B}^2$, and so $$\frac{OM}{OB}=\frac{OB}{OF}.$$ Now consider the triangles $BOM$ and $FOB$. They have a common angle and proportional adjacent sides, hence, they are similar. Let $$k=\frac{OM}{OB}=\frac{OB}{OF}=\frac{BM}{BF},$$ Fig. 32 Fig. 33

then $OB=kOF$, $OM=kOB=k^{2}OF$. This implies (see fig. 33) $$\frac{DM}{DF}=\frac{BO-OM}{OF-OB}=\frac{kOF-k^{2}OF}{OF-kOF}=k=\frac{BM}{BF},$$ and by the angle bisector theorem, $BD$ is the bisector of $\angle FBM$. This means that the angle between $BF$ and the bisector $BD$ equals to the angle between the bisector $BD$ and the median $BM$, and so $BF$ is indeed a "simedian" of the triangle $ABC$.

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Alternative solution.

Alternative solution. Consider the triangle $ACF$. The points $E$, $X$ and $B$ are collinear, so, by the Menelaus' theorem, we have $$\frac{AE}{EF}\cdot \frac{FB}{BC}\cdot \frac{CX}{XA}=1.(1)$$ Since $FX$ is the bisector of the angle $AFC$, we get $\frac{CX}{XA}=\frac{EF}{FB}$. $EA=EC$, as they are tangents to the same circle from the same point, and $BF=\frac{A{F}^2}{FC}$ from the secant-tangent theorem. So we can rewrite (1) in the form $$\frac{EC}{EF}\cdot \frac{AF}{BC}=1.(2)$$ Since $\angle ECA=\angle CBA$, we have $\angle FCE=\pi -\angle ECA-\angle ACB=\pi -\angle CBA-\angle ACB=\angle CAB$. So by the sine theorem for the triangle $CEF$ we get $$\frac{EC}{\sin \angle AFB}=\frac{EF}{\sin \angle FCE}=\frac{EF}{\sin \angle CAB},(3)$$ and, on the other hand, by the sine theorem for the triangle $AFB$ $$\frac{AF}{\sin \angle FBA}=\frac{AB}{\sin \angle AFB}.(4)$$ Using (2), (3), and (4) together with the sine theorem for the triangle $ABC$, we obtain $$\frac{AB}{BC}=\frac{EF\sin \angle AFB}{EC\sin \angle FBA}=\frac{\sin \angle CAB}{\sin \angle CBA}=\frac{BC}{AC},$$ which implies the necessary identity.