Solution II by A. Batominovski

Triangles with Equal Area II

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 is by Anton Batominovski.

First note that

Area(RPK) / Area(RQL) = RP · PK / RQ · QL

because ∠RPK = ∠RQL. Moreover, PK / QL = CP / CQ since ΔCPK and ΔCQL are similar. Therefore, PK / QL = PB / QA (since PK and QL are the perpendicular bisectors of BC and AC, respectively). Thus,

Area(RPK) / Area(RQL) = RP · PB / RQ · QA = Area(RPB) / Area(RQA) ,

since ∠BPR = ∠ACB = ∠AQR. However, BR = AR, ∠BPR = ∠AQR, and

	∠BRP	= ∠CAB
		= π - ∠ACB - ∠ABC
		= π - ∠AQR - ∠ARQ
		= ∠RQA

imply that the triangles RPB and RQA are equivalent. Hence,

Area(RPK) / Area(RQL) = Area(RPB) / Area(RQA) = 1,

and we are done.

2007 IMO, Problem 4

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