# 2002 JBMO Problems/Problem 2

## Problem

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 .

## Solution

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)

Now So,

giving .