## Triangles with Equal Area IV

Here is Problem 4 from the IMO 2007:

 In triangle ABC the bisector of angle BCA intersects the circumcircle again at R, the perpendicular bisector of BC at P, and the perpendicular bisector of AC at Q. The midpoint of BC is K and the midpoint of AC is L. Prove that the triangles RPK and RQL have the same area. The solution by Bui Quang Tuan is the nearest to the manner which is, to my taste, the most natural to treat this problem. Still, it requires additional construction. Let

• Ra, Rb be the orthogonal projections of R on CA, CB, respectively,
• Qa, Qb be the orthogonal projections of Q on RRa, RRb, respectively,
• Pa, Pb be the orthogonal projections of P on RRa, RRb, respectively,
• O' be the orthogonal projections of O on CR. ΔPOQ is isosceles because ∠OQP = ∠OPQ = 90° - ∠ACB /2 so PO' = QO', moreover O' is midpoint of chord CR, therefore QC = PR. It makes two right triangles CLQ and PPaR equal implying QL = RPa. As constructed QL||RPa therefore QLPaR is parallelogram. Similarly PKQbR is also parallelogram.

The rest is a proof without words:

 Area( ΔQLR) = Area( ΔQPaR) = Area( ΔPQaR) = Area( ΔPQbR) = Area( ΔPKR),

of which the only one that may require an explanation is Area( ΔQPaR) = Area( ΔPQaR) which is true because trianles PQPa and PQaPa have the same base (PPa) and lie between the same parallel lines and thus have the same area.

Note: also observes that the proof goes through for any P and Q on CR that are reflections of each other in the midpoint of CR. ### 2007 IMO, Problem 4 