The Plemelj Construction of a Triangle: 4
This Demonstration constructs a triangle given the length of its base , the length of the altitude from to and the difference between the angles at and .[more]
Step 1: Draw a line segment of length and its midpoint .
Step 2: Draw a line segment of length perpendicular to . Draw the line through parallel to .
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 and . 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 , (Euclid, Book 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, .[less]
For the history of this problem, references and a photograph of Plemelj's first solution, see The Plemelj Construction of a Triangle: 1.
 Euclid, Elements, Vol. 1, New York: Dover Publications, 1956 pp. 311–312.