MathLabs

Problem 1

Let ABCABC be a triangle, and let DD be a point on side BCBC. A line through DD intersects side ABAB at XX and ray ACAC at YY. The circumcircle of triangle BXDBXD intersects the circumcircle ω\omega of triangle ABCABC again at point Z≠BZ\ne B. The lines ZDZD and ZYZY intersect ω\omega again at VV and WW, respectively. Prove that AB=VWAB=VW.
Step 1 of 3: Create the first angle chain
B,X,D,Z cyclic,A,B,C,Z,V cyclicB,X,D,Z\text{ cyclic},\quad A,B,C,Z,V\text{ cyclic}
Detailed analysis

The points B,X,D,Z are concyclic by construction, and A,B,C,Z,V lie on omega. Since D,Y,X are collinear, cyclic angles give angle ZDY equal to angle ZBA, while A,B,C,Z being cyclic gives angle ZBA equal to angle ZCY.