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Problem 192 from Connelly's corpus: Ea, Eb, Hc


Problem

Given a point Ea, a point Eb and a point Hc, 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 Ea, construct a circle k(Ea,A) (rule W06);

    % NDG: points Hc and Ea are not the same

  2. 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

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

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

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

  4. Using the point H and the point Ea, construct a point A (rule W01);

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

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

    % DET: points Eb and H are not the same

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

    % DET: points Hc and H are not the same

  8. Using the circle k(Ea,A), the line hb, the point Ea and the point H, construct a point Hb (rule W05);

    % NDG: line hb and circle k(Ea,A) intersect

    % DET: points H and Hb must be different

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

    % DET: points A and Hb are not the same

  10. Using the line hc and the line b, construct a point C (rule W03);

    % NDG: lines hc and b are not parallel

    % DET: lines hc and b are not the same

animation
Construction in GCLC language

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