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

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This Demonstration shows an alternative 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.

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Construction

Step 1: Draw the line segment of length and the line segment of length perpendicular to at . Extend to so that as well.

Step 2: Draw a circle with center subtending the central angle over the chord so that .

Let the point be the intersection of the circle and the line through parallel to .

Step 3: Let be the point on such that is the midpoint of . Construct the point on the ray so that .

Step 4: Let be the intersection of and .

Then satisfies the conditions.

Verification

Let be the intersection of and .

The distance of to is . The segment bisects the angle at since it is the right bisector of .

Since the triangles and are congruent, the quadrilateral is a rhombus. Since , , the angle at is because the lines and are parallel.

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Contributed by: Nada Razpet, Marko Razpet and Izidor Hafner (August 2017)
Open content licensed under CC BY-NC-SA


Snapshots


Details

The problem appeared in Wiegand's book [4, p. 147]; the photograph is from that book. One solution is also given in [5, p.156].

When professor Plemelj was in Chernivtsi (1908–1914, then in the Austrian–Hungarian Empire, now in Ukraine), he talked with two students about the construction problem in a certain textbook that he had been given by his teacher Borstner when attending gymnasium. They brought him a copy of the book, but Plemelj forgot for the second time to write down the title, so we still do not know its title.

He looked for the book in Ljubljana after the First World War but could not find it, but he found the problem of this Demonstration in [4].

Using the construction from [4], Plemelj found a nice solution of his original problem (given base , the altitude and the difference of the angles at base).

This Demonstation does the opposite. Using the construction from Borstner’s book, we solve the problem in Wiegand’s book. An almost identical approach is given in [5, p. 93].

References

[1] Wikipedia. "Josip Plemelj." (Aug 23, 2017) en.wikipedia.org/wiki/Josip_Plemelj.

[2] J. J. O'Connor and E. F. Robertson. "Josip Plemelj." MacTutor. (Aug 23, 2017) www-history.mcs.st-andrews.ac.uk/Biographies/Plemelj.html.

[3] J. Plemelj, Iz mojega življenja in dela (From My Life and Work), Obzornik mat. fiz., 39, 1992 pp. 188–192.

[4] M. Bland and A. Wiegand, Geometrische Aufgaben für Hohëre Lehranstalten, Braunschweig: Schewetschke und Sohn, 1865.

[5] D.S. Modic, Trikotniki, Konstrukcije, Algebrske Rešitve, Ljubljana: Math d.o.o., 2009.



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