Back to the list of problems

Problem 441 from Connelly's corpus: Ec, Ha, Mb


Problem

Given a point Ec, a point Ha and a point Mb, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


  1. Using the point Ha and the point Ec, construct a circle k(Ec,C) (rule W06);

    % NDG: points Ha and Ec are not the same

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

    % NDG: points Ha and Mb are not the same

  3. Using the circle k(Ec,C), the circle k(Mb,C), the point Ha, the point Ec and the point Mb, construct a point C (rule W08);

    % NDG: circles k(Ec,C) and k(Mb,C) intersect

    % DET: circles k(Ec,C) and k(Mb,C) are not the same points Ha and C must be different

  4. Using the point C and the point Ec, construct a point H (rule W01);

  5. Using the point Mb and the point C, construct a point A (rule W01);

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

    % DET: points Ha and C are not the same

  7. Using the point Mb and the point C, construct a line b (rule W02);

    % DET: points Mb and C are not the same

  8. Using the circle k(Ec,C), the line b, the point Ec and the point C, construct a point Hb (rule W05);

    % NDG: line b and circle k(Ec,C) intersect

    % DET: points C and Hb must be different

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

    % DET: points H and Hb are not the same

  10. Using the line a and the line hb, construct a point B (rule W03);

    % NDG: lines a and hb are not parallel

    % DET: lines a and hb are not the same

animation
Construction in GCLC language

Back to the list of problems