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 . This is not Plemelj's construction, but a new one that unifies Plemelj's first construction and an alternative one.
Step 1: Draw a straight line of length and a perpendicular line segment with midpoint .
Step 2: Draw a circle with center such that is viewed at an angle from points on below the chord . Let be the midpoint of . The angle equals .
Step 3: Find a point on the circle at distance from and a point at distance from .
Step 4: Draw the isosceles trapezoid .
Step 5: The point is the intersection of the straight line through parallel to and the right bisector of and .
Step 6: The triangle meets the stated conditions.
This is similar to Plemelj's first construction, but instead of triangle , start with triangle , which is also congruent to . In the isosceles triangle , , so . The obtuse angle ; .