Comment: It comes as a pleasant surprise to me that in Western part of China they do know about Apollonious theorem concerned with a tangent to hyperbola! Thank you (Chairman Mao?). M.T.

I solved it,but I use at least 4 hours!!! Oh,my God.

What's that?? I mean, stealing problems from other olympiads is OK if one has no phantasy; proposing ugly but original problems is also halfways OK, as a matter of honor, but stealing ugly problems from other sources sux, particularly if the other source is the ISL 1998. To make it short: The problem is equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 . Finding out why is not that easy, since the proposers at the CWMO apparently tried everything to make the situation as confusing as possible, so here are the details: First let me rewrite the problem in a constructive way: Problem. Let $k_1$ be a circle with center $O_1$, and let D and A be two points on this circle. The tangent to the circle $k_1$ at the point A intersects the line $O_1D$ at a point C. The perpendicular to the line $O_1D$ through the point A meets the circle $k_1$ at a point E (apart from A). Let F be the feet of the perpendicular from the point A to the line DE. Let $k_2$ be the circle through the points A and C which touches the line $O_1D$ at the point C. (You can easily see that there is only one such circle: The center of this circle must lie on the perpendicular bisector of the segment AC - since the segment AC is a chord of this circle - and on the perpendicular to the line $O_1D$ at the point C - since the circle touches the line $O_1D$ at the point C; hence, the center of this circle is uniquely determined as the point of intersection of these two lines; moreover, since we know a point on the circle - namely, the point A -, the radius is uniquely determined, too, so there is only one such circle.) Let B be the point of intersection of the circles $k_1$ and $k_2$ (different from A). Prove that the line BD bisects the segment AF. Solution. Let Y be the midpoint of the segment AF. Then, we want to show that the line BD passes through this point Y. Let G be the point diametrically opposite to the point D on the circle $k_1$, i. e. the reflection of the point D in the center $O_1$ of this circle. Then, the segment DG is a diameter of the circle $k_1$, and thus < DAG = 90°. Since the segment AE is a chord of the circle $k_1$, the perpendicular bisector of this segment AE must pass through the center $O_1$ of this circle $k_1$. In other words, the perpendicular bisector of the segment AE is the perpendicular to the line AE through the point $O_1$. But the perpendicular to the line AE through the point $O_1$ is the line DG (since the line DG coincides with the line $O_1D$, and $AE\perp O_1D$). Hence, the perpendicular bisector of the segment AE is the line DG. In other words, the point E is the reflection of the point A in the line DG. Now, let the line DY intersect the circle $k_1$ at a point B' (apart from D). In order to solve the problem, it will be enough to prove that B' = B. In fact, the point B' is known to lie on the line DY; in other words, the line B'D passes through the point Y; hence, once it will be shown that B' = B, we will conclude that the line BD passes through the point Y, and the problem will be solved. Now, consider our situation from another viewpoint: We have a triangle ADG with < DAG = 90°. Its circumcircle is $k_1$, and the tangent at the point A to this circumcircle $k_1$ intersects the line DG at the point C. The point E is the reflection of the point A in the line DG. The point F is the foot of the perpendicular from the point A to the line DE. The point Y is the midpoint of the segment AF. The line DY intersects the circle $k_1$ at a point B' (apart from D). Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=18215 , the line DC (i. e., the line $O_1D$) is tangent to the circumcircle of triangle ACB'. Thus, the circumcircle of triangle ACB' is the circle through the points A and C which touches the line $O_1D$ at the point C. Hence, the circumcircle of triangle ACB' is simply the circle $k_2$. Consequently, the point B' lies on the circle $k_2$. Thus, the point B' is the point of intersection of the circles $k_1$ and $k_2$ (different from A), and this yields B' = B, so the problem is solved. Darij

My solution: Let $M=AE \cap CD , G=AF \cap BD$ . Easy to see $M$ is the midpoint of $AE$ . Since $\angle BCM=\angle BAC=\angle BEM$ , so we get $B, C, E, M$ are concyclic , hence $\angle BMA=90-\angle CMB=90-\angle CEB=90-\angle EDB=\angle BGA$ . ie. $A, B, G, M$ are concyclic Since $\angle AMG=\angle ABD=\angle AED$ , so we get $MG \parallel DE$ . ie. $AG=GF$ Q.E.D