First, note that since X lies on the circumcircle of triangle WXY, we have angle(WXY) = angle(XZY), where ZY is a tangent to the circumcircle at X. Also, since WZ is parallel to ZY, we have angle(WZX) = angle(XZY).
Using the fact that the sum of angles in a triangle is 180 degrees, we can write:
angle(WXY) + angle(WZX) + angle(XZY) = 180 degrees
Substituting angle(WXY) = angle(XZY), we get:
2 angle(XZY) + angle(WZX) = 180 degrees
Since angle(WZX) = angle(XZY), we have:
3 angle(XZY) = 180 degrees
So, angle(XZY) = 60 degrees.
Now, consider triangle WXY. Using the law of cosines, we have:
XY^2 = WX^2 + WY^2 - 2 WX WY cos(angle(WXY))
Substituting XY = 14, WX = 6, and angle(WXY) = 120 degrees (since angle(WXY) + angle(XWY) + angle(XYW) = 180 degrees), we get:
14^2 = 6^2 + WY^2 - 2 * 6 * WY * (-1/2)
Simplifying, we get:
WY^2 + 6 WY - 40 = 0
Factoring, we get:
(WY - 4)(WY + 10) = 0
Since WY cannot be negative, we have WY = 4.
Then from similar triangles XWZ and YWX, YZ = 4*6/14 = 12/7.
