25. Construct a Triangle Given Its Base, the Difference of the Base Angles and the Length of One of Three Line Segments

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This Demonstration constructs a triangle given the length of its base , the difference between the angles and of the base and the length of one of the following:


i. the altitude from to the base

ii. the bisector from to the base

iii. the line segment through the circumcenter from to the base

The last two cases are reduced to the first case, since the angle between the bisector and the altitude is , and the angle between the altitude and the line through the center is . So we have a Plemelj triangle, for which we use The Plemelj Construction of a Triangle: 13.


1. Draw line of length .

2. Draw a circle of radius with center , the midpoint of .

3. Draw a ray from at an angle to intersect at the point . Let be a point such that the length of is the length of the given altitude and is perpendicular to .

4. Draw a circle with center and radius . Let the point be on the circle such that . The point is the intersection of the perpendicular bisector of and the line parallel to through .


The angles and are equal to . The angle at is .

The first case was posed to Josip Plemelj (1873–1967) by his mathematics teacher in secondary school in 1891 (in Ljubljana, then in the Austro-Hungarian Empire, now in Slovenia).


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



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