parmenides51 wrote: Two different points $X,Y$ are marked on the side $AB$ of a triangle $ABC$ so that $\frac{AX \cdot BX}{CX^2}=\frac{AY \cdot BY}{CY^2}$ . Prove that $\angle ACX=\angle BCX$. I.Zhuk Is prove that $\angle ACX=\angle BCY$.

$\angle ACY=\alpha$ , $\angle XCB=\beta$, $\angle YCX=\theta$, $\angle BAC=\omega$ $$\frac{sin(\alpha+\theta).sin(\beta)}{sin(\omega)sin(\alpha+\theta+\beta+\omega)}=\frac{AX \cdot BX}{CX^2}=\frac{AY \cdot BY}{CY^2}=\frac{sin(\theta+\beta).sin(\alpha)}{sin(\omega)sin(\alpha+\theta+\beta+\omega)}$$$$sin(\alpha+\theta).sin(\beta)=sin(\theta+\beta).sin(\alpha)$$$$cos(\alpha+\theta-\beta)-cos(\alpha+\theta+\beta)=cos(\theta+\beta-\alpha)-cos(\alpha+\theta+\beta)\Rightarrow cos(\alpha+\theta-\beta)=cos(\theta+\beta-\alpha)$$$$2sin(\theta)sin(\beta-\alpha)=0\Rightarrow \beta=\alpha\Rightarrow \angle ACY=\angle BCX \Rightarrow \angle ACX=\angle BCY$$

By the Forgotten Coaxiality Lemma, the circles $(CAB), (CXY)$ and the point circle $C$ are coaxial, so $(CXY)$ and $(CAB)$ are tangent at $C$, Now the inversion at $C$ with radius $\sqrt{CA\cdot CB}$ composed with a reflection about the bisector of $\angle C$ takes $X,Y$ to $X^*,Y^*$, two points on $\widehat{BA}$ with $X^*Y^*||AB$. Thus $X^*Y^*BA$ is an isosceles trapezium, so $\widehat{X^*A}=\widehat{Y^*B}$, implying the result.

Let the line $CX$ meet the circumcircle $\omega$ of $\triangle ABC$ at $R$, and let the line $CY$ meet $\omega$ at $S$. Then $AX \cdot BX = CX \cdot XR \implies \frac{AX \cdot BX}{CX^2} = \frac{XR}{CX}$. Similarly, $\frac{AY \cdot BY}{CY^2} = \frac{YS}{CY}$. Thus, $\frac{XR}{CX} = \frac{YS}{CY} \implies XY \parallel RS$, so arcs $AR$ and $BS$ of $\omega$ must be equal as parallel chords cut off equal arcs. Thus, $\angle ACX = \angle ACR = \angle BCS = \angle BCY$.