Here is a nice figure. We give a hint for the solution and invite you to give the details

Let ABC be a triangle.
Let AA1, BB1, CC1 be its three bisectors.

Let A2, B2 and C2 be the intersections between the bisector and the circumcircle.

Let A3, B3 and C3 be the projections of the innercircle on the three sides of the triangle.

Then A2A3, B2B3 and C2C3 intersect.

Solution is simple and we want to give you a hint inviting you to find out why there are parallel sides of two triangles in the next figure