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SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Saudi Arabia geometry

Problem

Let be a non-isosceles triangle inscribed in a circle and , are two angle bisectors intersecting at with belonging to segment and belonging to segment . Suppose that , intersect at , respectively. The line passes through and is perpendicular to , intersecting at the second point ; the line passes through and is perpendicular to , intersecting at the second point . Denote , as the midpoints of and respectively.

1. Prove that triangles and are similar.

2. Suppose that is the intersection of the two lines and . Prove that is perpendicular to .

problem
Solution
1) Denote the projections of on , are , respectively. Note that , , are collinear because . Because , are the midpoints of , respectively, we have and . Similarly, and . Hence, . On the other hand, So implies that .



2) It is easy to see that with the radius of . Then or , , , are cyclic. From this, we can see so . The antipole line of , passes through so the antipole line of is , which leads to . From these results, we get .

Techniques

Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleTrigonometryCyclic quadrilateralsPolar triangles, harmonic conjugatesInversionAngle chasing