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Problem 137 from Connelly's corpus: C, Ea, Ma


Problem

Given a point C, a point Ea 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 C and the point Ma, construct a point B (rule W01);

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

    % DET: points C and Ma are not the same

  3. Using the point Ea and the point Ma, construct a line m(HbHc) (rule W02);

    % DET: points Ea and Ma are not the same

  4. Using the point Ea and the line a, construct a line ha (rule W10b);

  5. Using the line ha and the line a, construct a point Ha (rule W03);

    % NDG: lines ha and a are not parallel

    % DET: lines ha and a are not the same

  6. Using the point C and the point Ha, construct a line m(CHa) (rule W14);

    % DET: points C and Ha are not the same

  7. Using the point Ea and the point Ma, construct a line m(EaMa) (rule W14);

    % DET: points Ea and Ma are not the same

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

    % NDG: lines m(EaMa) and m(HbHc) are not parallel

    % DET: lines m(EaMa) and m(HbHc) are not the same

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

    % NDG: points Ea and N are not the same

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

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

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

animation
Construction in GCLC language

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