Two circles intersects in X and Y. The tangent at X to the first circle cuts the second circle at A and AY produced cuts the first circle at B. Prove that XB is parallel to the tangent at A to the second circle.
Extend $\displaystyle AX $ we have $\displaystyle X$ being in the segment $\displaystyle AM$ therefore , $\displaystyle \angle MXB = \angle BYX$ Moreover , $\displaystyle \angle NAX = \angle AYX $ where $\displaystyle AN $ is the tagent to the second circle and $\displaystyle N $ and $\displaystyle B $ are in the opposite side .
By looking at line $\displaystyle AYB $ we obtain
$\displaystyle 180^o = \angle BYX + \angle AYX = \angle MXB + \angle NAX $
$\displaystyle \angle AXB = \angle NAX $