2011 USAJMO Problems/Problem 5
Points , , , , lie on a circle and point lies outside the circle. The given points are such that (i) lines and are tangent to , (ii) , , are collinear, and (iii) . Prove that bisects .
Since , we have that . But it is well known that is a harmonic quadrilateral, thus is a symmedian of triangle , from which it follows that is a median of .
Connect segment PO, and name the interaction of PO and the circle as point M.
Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.
∠ BOA = 1/2 arc AB + 1/2 arc CE
Since AC // DE, arc AD = arc CE,
thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM
Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)
BE bisects AC, proof completed!
~ MVP Harry
Let be the center of the circle, and let be the intersection of and . Let be and be .
Thus is a cyclic quadrilateral and and so is the midpoint of chord .
This is the solution from EGMO Problem 1.43 page 242
Let be the center of the circle, and let be the midpoint of . Let denote the circle with diameter . Since , , , and all lie on .
Since quadrilateral is cyclic, . Triangles and are congruent, so , so . Because and are parallel, lies on (using Euclid's Parallel Postulate).
-Evan Chen (vEnhance)
Note that by Lemma 9.9 of EGMO, is a harmonic bundle. We project through onto , Where is the point at infinity for parallel lines and . Thus, we get , and is the midpoint of . ~novus677