This Demonstration constructs a triangle given the length of its base, the difference of the base angles and a line (given by a point and slope ) that contains . This generalizes the problem given to Plemelj when he was in high school, namely, the problem in which the line is parallel to . Our construction is a simple adaptation of The Plemelj Construction of a Triangle: 4. Construction Step 1: Draw a line segment of length and its midpoint . Draw a line . Step 2: Draw a line segment from that is perpendicular to . It meets the line at . Step 3: Choose any point on the ray . Step 4: Let be on such that . Step 5: Let be the circle with center and radius . Step 6: Let be the intersection of and the ray . Join and . Step 7: The point is the intersection of and the line through parallel to . Step 8: The triangle meets the stated conditions. Theorem: Let be any triangle, and let be the foot of the altitude from to . Let be the circumcenter of , and let the segment be perpendicular to with on the same side of as . Then . Proof: The inscribed angle subtended by the chord is , so the central angle . Since is isosceles, . Since , . Since , [1, Proposition 29]. So . Now in the construction, let be the intersection of and the line through parallel to . Then the isosceles triangles and are similar, and is parallel to , so . By construction, , so . Since is on the right bisector of , it is the circumcenter of . By the theorem, .
[1] Euclid, Elements, Vol. 1, 2nd ed. (T. Heath, ed.), New York: Dover Publications, 1956 pp. 311–312. [2] E. Heis and T. J. Eschweiler, Lehrbuch der Geometrie zum Gebrauche an höheren Lehranstalten, Köln: Verlag der M. DumontSchauberg'schen Buchhandlung, 1855.
