Problem 2 from the IMO 2007, Second solution
Here is Problem 2 from the IMO 2007:
Consider five points A, B, C, D and E such that ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let i be a line passing through A. Suppose that l intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that
The solution below is by Vo Duc Dien and is a significant simplification of the solution by Anton Batominovski.
Based on Simson-Wallace's theorem the feet of projections of E down to the three sides of triangle BCD (denoted M, L and K) are colinear as seen on the diagram.
Since L and K are the midpoints of FC and CG, respectively, LK||FG, and in triangle ACG, LK intersects AC at its midpoint M. Since ABCD is a parallelogram, M is also the midpoint of DB.
Therefore, there is only one unique point M to satisfy conditions that M is on DB and also collinear with K and L, and M is also the foot of E to DB.
Extend EM to cut the circle at I. Since EI is perpendicular to DB and
It follows that ∠ECI = 90°. Further,
|(*)||∠4 = ∠5.|
|In triangle EFL:||∠1 + ∠2 + ∠3 = 90°|
|In triangle EFG:||2(∠1 + ∠2 + ∠5) = 180°|
|or||∠1 + ∠2 + ∠5 = 90°.|
|Therefore||∠3 = ∠5.|
From (*), ∠3 = ∠4. In other words,