MathDB
Slovenia 2019 TST1 P5

Source: 2019 Slovenia 1st TST Problem 5

February 19, 2019
TSTgeometry

Problem Statement

Let ABCABC be a triangle and D,ED, E and FF the foots of heights from A,BA, B and CC respectively. Let D1D_1 be such a point on EFEF, that DF=D1EDF = D_1 E where EE is between D1D_1 and FF. Similarly, let D2D_2 be such a point on EFEF, that DE=D2FDE = D_2 F where FF is between EE and D2D_2. Let the bisector of DD1DD_1 intersect ABAB at PP and let the bisector of DD2DD_2 intersect ACAC at QQ. Prove that, PQPQ bisects BCBC.