Given a parallelogram \(ABCD\). Let \(\ell_1\) be the line through \(D\), parallel to \(AC\), and \(\ell_2\) the external bisector of \(\angle ACD\). The lines \(\ell_1\) and \(\ell_2\) intersect at \(E\). The lines \(\ell_1\) and \(AB\) intersect at \(F\), and the line \(\ell_2\) intersects the internal bisector of \(\angle BAC\) at \(X\). The line \(BX\) intersects the circumcircle of triangle \(EFX\) at a second point \(Y\). The internal bisector of \(\angle ACD\) intersects the circumcircle of triangle \(ACX\) at a second point \(Z\). Prove that the quadrilateral \(DXYZ\) is inscribed in a circle.
Problem
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
Tags: geometry
Assassino9931
10.09.2024 21:15
Proposed by @Marinchoo.
trigadd123
14.11.2024 01:42
Very nice problem! We prove more strongly that $DXYZ$ is an isosceles trapezoid. First notice that $\angle AXC=90^{\circ}$. Furthermore $CZ\parallel AX$, so $AXCZ$ is a rectangle. Therefore $XZ=AC$. We now show that $BX\parallel DZ$. This follows because $X$ and $Z$ are reflections across the midpoint of $AC$ and as are $B$ and $D$, so $DZBX$ is a parallelogram. Therefore the desired reduces to showing that $DY=XZ$, which is equivalent to showing that $DY=AC$. The key insight is constructing $K$, the intersection of $BX$ and $CD$. Claim.$DK=AC$. Proof. Since $CK$ is the external angle bisector of $\angle ACD$, it is clear that $L$ is the reflection of $A$ across $X$. Now $KD=AC$ is equivalent to $KD=CL$, which reduces to showing that $KL=CD$. But $AB\parallel CL$ and $AX=XK$, so $KL=AB=CD$, as needed. Observe that $DFAC$ is a parallelogram, so $DF=AC$. It thus suffices to show that $D$ is the circumcenter of $\triangle FKY$. Since $DK=DF$, this reduces to proving that $\angle FDK=2\angle FYK$. But $\angle FYK=\angle DEC$ and $\angle FDK=180^{\circ}-\angle EDC$. The desired easily follows since $DE\parallel AC$ and $CE$ is the external bisector of $\angle ACD$.