2002 JBMO Problems/Problem 2
Two circles with centers and meet at two points and such that the centers of the circles are on opposite sides of the line . The lines and meet their respective circles again at and . Let be the midpoint of . Let , be points on the circles of centers and respectively, such that , and lies on the minor arc while lies on the minor arc . Show that .
It's easy to see that forms a straight line and is parallel to line .
Let us define .
Let circumradii of the 2 circles be and respectively.
Now and , this implies that:
. So forms a straight line.
Now since is the midpoint of , is parallel to and its length is equal to .
Similarly, we see that, is parallel to and it's length is equal to .
So (since is the circumcenter).
So forms a cyclic quadrilateral.
Thus, we have .
Adding to both sides we have: or,
Thus by SAS, is congruent to
So, we have , hence is an isoceles triangle.
So, we get -- (1)