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Problem 368 from Connelly's corpus: Eb, Hc, Ma


Problem

Given a point Eb, a point Hc and a point Ma, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


  1. Using the point Hc and the point Eb, construct a circle k(Eb,B) (rule W06);

    % NDG: points Hc and Eb are not the same

  2. Using the point Hc and the point Ma, construct a circle k(Ma,B) (rule W06);

    % NDG: points Hc and Ma are not the same

  3. Using the circle k(Eb,B), the circle k(Ma,B), the point Hc, the point Eb and the point Ma, construct a point B (rule W08);

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

    % DET: circles k(Eb,B) and k(Ma,B) are not the same points Hc and B must be different

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

  5. Using the point Ma and the point B, construct a point C (rule W01);

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

    % DET: points Hc and B are not the same

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

    % DET: points Ma and B are not the same

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

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

    % DET: points B and Ha must be different

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

    % DET: points H and Ha are not the same

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

    % NDG: lines c and ha are not parallel

    % DET: lines c and ha are not the same

animation
Construction in GCLC language

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