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Solution of a Geometrical Problem

1903
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Proceedings of the Edinburgh Mathematical Society
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To draw through a given point a transversal of a given triangle so that the segments of the transversal may be in a given ratio." FIGURE 25. Analysis. ABC is a triangle and DEF a transversal and K is the point of concurrence of the four circles circumscribed about the four triangles formed by the transversal and the sides of the triangle. H is a point on the circumference of the circumcircle of ABC, such that AK and AH are equally inclined to the bisector of the angle BAC. Thus z.HBC = . L K C

doi:10.1017/s0013091500001991
fatcat:d263e6x2xfdvjcux7l74uvipoi