## Triangles with Equal Area V

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. |

This is another solution by Bui Quang Tuan. Suppose _{1}, RD at Q'. A line through P and perpendicular to AC (RE) intersects AC at K_{1}, RE at P'. By symmetries of rhombus in center O',

Area(RPK) | = Area(RQL_{1}K) - Area(PQL_{1}K) | |

= Area(RQLK_{1}) - Area(KL_{1}Q'K_{2})/2 | ||

= Area(RQLK_{1}) - Area (LK_{1}P'L_{2})/2 | ||

= Area(RQLK_{1}) - Area(LK_{1}R) | ||

= Area(RQL). |

### 2007 IMO, Problem 4

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny65100832 |