 # The Plemelj Construction of a Triangle: 4

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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 .

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Construction

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.

Verification

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, .

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

## Snapshots   ## Details

For the history of this problem, references and a photograph of Plemelj's first solution, see The Plemelj Construction of a Triangle: 1.

Reference

 Euclid, Elements, Vol. 1, New York: Dover Publications, 1956 pp. 311–312.

## Permanent Citation

Izidor Hafner

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