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Problem 246 from Connelly's corpus: Ea, Ha, Mc


Problem

Given a point Ea, a point Ha 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 Ea and the point Ha, construct a line ha (rule W02);

    % DET: points Ea and Ha are not the same

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

    % NDG: points Ha and Mc are not the same

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

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

    % DET: points Ha and A must be different

  4. Using the point A and the point Ea, construct a point H (rule W01);

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

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

    % DET: points Ha and B are not the same

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

    % DET: points H 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 A and the point Hb, construct a line b (rule W02);

    % DET: points A and Hb 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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