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Problem 357 from Connelly's corpus: Eb, Hb, Hc


Problem

Given a point Eb, a point Hb 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 Eb and the point Hb, construct a line hb (rule W02);

    % DET: points Eb and Hb 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(Eb,B) and the line hb, construct a point B and a point H (rule W04);

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

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

    % DET: points Hc and B are not the same

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

    % DET: points Hc and H are not the same

  6. Using the point Hb and the line hb, construct a line b (rule W10a);

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

    % NDG: lines b and c are not parallel

    % DET: lines b and c are not the same

  8. 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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