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Problem 110 from Connelly's corpus: B, Ec, Mc


Problem

Given a point B, a point Ec 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 B and the point Mc, construct a point A (rule W01);

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

    % DET: points B and Mc are not the same

  3. Using the point Ec and the point Mc, construct a line m(HbHa) (rule W02);

    % DET: points Ec and Mc are not the same

  4. Using the point Ec and the line c, construct a line hc (rule W10b);

  5. Using the line hc and the line c, construct a point Hc (rule W03);

    % NDG: lines hc and c are not parallel

    % DET: lines hc and c are not the same

  6. Using the point B and the point Hc, construct a line m(BHc) (rule W14);

    % DET: points B and Hc are not the same

  7. Using the point Ec and the point Mc, construct a line m(EcMc) (rule W14);

    % DET: points Ec and Mc are not the same

  8. Using the line m(EcMc) and the line m(HbHa), construct a point N (rule W03);

    % NDG: lines m(EcMc) and m(HbHa) are not parallel

    % DET: lines m(EcMc) and m(HbHa) are not the same

  9. Using the point Ec and the point N, construct a circle k(N,Ma) (rule W06);

    % NDG: points Ec and N are not the same

  10. Using the circle k(N,Ma) and the line m(BHc), construct a point Ma and a point Eb (rule W04);

    % NDG: line m(BHc) and circle k(N,Ma) intersect

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

animation
Construction in GCLC language

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