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Problem 309 from Connelly's corpus: Eb, Ec, Ha


Problem

Given a point Eb, a point Ec and a point Ha, 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 Eb, construct a circle k(Eb,B) (rule W06);

    % NDG: points Ha and Eb are not the same

  2. 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

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

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

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

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

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

  6. Using the point Ec and the point H, construct a line hc (rule W02);

    % DET: points Ec and H are not the same

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

    % DET: points Ha and H are not the same

  8. Using the circle k(Eb,B), the line hc, the point Eb and the point H, construct a point Hc (rule W05);

    % NDG: line hc and circle k(Eb,B) intersect

    % DET: points H and Hc must be different

  9. Using the point B and the point Hc, construct a line c (rule W02);

    % DET: points B and Hc are not the same

  10. Using the line ha and the line c, construct a point A (rule W03);

    % NDG: lines ha and c are not parallel

    % DET: lines ha and c are not the same

animation
Construction in GCLC language

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