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Problem 34 from Connelly's corpus: A, Eb, Mc


Problem

Given a point A, a point Eb and a point Mc, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


  1. Using the point A and the point Mc, construct a point B (rule W01);

  2. Using the point Eb and the point B, construct a point H (rule W01);

  3. Using the point A and the point H, construct a line ha (rule W02);

    % DET: points A and H are not the same

  4. Using the point Eb and the point B, construct a line hb (rule W02);

    % DET: points Eb and B are not the same

  5. Using the point A and the point Mc, construct a circle k(Mc,A) (rule W06);

    % NDG: points A and Mc are not the same

  6. Using the circle k(Mc,A), the line ha, the point Mc and the point A, construct a point Ha (rule W05);

    % NDG: line ha and circle k(Mc,A) intersect

    % DET: points A and Ha must be different

  7. Using the point Ha and the point B, construct a line a (rule W02);

    % DET: points Ha and B are not the same

  8. Using the circle k(Mc,A), the line hb, the point Mc and the point B, construct a point Hb (rule W05);

    % NDG: line hb and circle k(Mc,A) intersect

    % DET: points B and Hb must be different

  9. Using the point Hb and the point A, construct a line b (rule W02);

    % DET: points Hb and A are not the same

  10. Using the line a and the line b, construct a point C (rule W03);

    % NDG: lines a and b are not parallel

    % DET: lines a and b are not the same

animation
Construction in GCLC language

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