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Perpendicular lines, equal angles and a fixed point

Source: Bundeswettbewerb Mathematik 2018, Round 2, Problem 3

February 5, 2019
geometrygeometry unsolvedperpendicular linesLocus

Problem Statement

Let TT be a point on a line segment ABAB such that TT is closer to BB than to AA. Show that for each point C≠TC \ne T on the line through TT perpendicular to ABAB there is exactly one point DD on the line segment ACAC with ∠CBD=∠BAC\angle CBD=\angle BAC. Moreover, show that the line through DD perpendicular to ACAC intersects the line ABAB in a point EE which is independent of the position of CC.
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