26c. Construct a Triangle Given the Length of Its Base, the Angle Opposite the Base and the Length of That Angle's Bisector

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This Demonstration constructs a triangle given the length of its base , the angle at the point , and the length of the angle bisector at . The Demonstration uses the conchoid of Nicomedes, which is shown in red.

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Draw a right-angled triangle with hypotenuse and . Let be on such that . Let and be on the perpendicular to at on either side of such that and .

Define the conchoid determined by and .

Let be the line from that forms an angle with the horizontal line . Let be the foot of the perpendicular from to .

Let be the intersection of and the conchoid. Let be the intersection of and . Then triangle is the solution.

By definition of the conchoid determined by the base with pole and the number , . By construction, . Since and are perpendicular projections of of the same size, is the angle bisector at and by construction .

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


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Details

The base of a conchoid is a straight line . Let (the pole of the conchoid) be a point not on the base such that the distance of from is . Let be a ray from not parallel to the base. The ray intersects the base at a point . Measure out points and on the ray so that , where is a positive number. The conchoid determined by the base, the pole and is the set of all such points and .



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