17a. Construct a Triangle Given the Length of an Angle Bisector, the Angle at Another Vertex and Distance from the Third Vertex to the Angle Bisector

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This Demonstration shows a construction of a triangle given the angle at , the length of the angle bisector at and the length of the perpendicular from to the bisector.



Step 1: Draw a line segment of length ; will turn out to be on . Construct a point such that the triangle is isosceles (with ) such that the angle at is . Construct a circle with center and radius .

Step 2: Draw a line segment of length perpendicular to at . Let be the midpoint of . Let be the intersection of ray and the circle. Let and be the intersections of and with a line perpendicular to at , respectively.

Step 3: Let be the point symmetric to with respect to , and let be the midpoint of , which is also on .

Step 4: Then is a solution of the problem.


The angle at is since it is an inscribed angle over the chord , and the central angle is .

Triangles and are congruent, so , and are collinear and . So is the bisector of the angle at of length . Finally, .


Contributed by: Izidor Hafner (August 2017)
Open content licensed under CC BY-NC-SA



This is problem 15 from [1, pp. 147].


[1] M. Bland and A. Wiegand, Geometrische Aufgaben für höhere Lehranstalten, Braunschweig: Schwetschke und Sohn, 1865. babel.hathitrust.org/cgi/pt?id=njp .32101076800687;view=1up;seq=95.

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