20b. Construct a Triangle Given the Length of Its Base, the Difference of Its Base Angles and a Line Containing the Third Vertex
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.
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, .