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Problem 450 from Connelly's corpus: Ec, Hb, Ma


Problem

Given a point Ec, a point Hb 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 Hb and the point Ec, construct a circle k(Ec,C) (rule W06);

    % NDG: points Hb and Ec are not the same

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

    % NDG: points Hb and Ma are not the same

  3. Using the circle k(Ec,C), the circle k(Ma,B), the point Hb, the point Ec and the point Ma, construct a point C (rule W08);

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

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

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

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

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

    % DET: points Hb and C are not the same

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

    % DET: points Ma and C are not the same

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

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

    % DET: points C 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 b and the line ha, construct a point A (rule W03);

    % NDG: lines b and ha are not parallel

    % DET: lines b and ha are not the same

animation
Construction in GCLC language

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