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Problem 241 from Connelly's corpus: Ea, Ha, Hb


Problem

Given a point Ea, a point Ha and a point Hb, construct the triangle ABC.

Status

ArgoTriCS says that the problem is solvable.

Illustration

Figure of construction

Solution


  1. Using the point Ea and the point Ha, construct a line ha (rule W02);

    % DET: points Ea and Ha are not the same

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

    % NDG: points Hb and Ea are not the same

  3. Using the circle k(Ea,A) and the line ha, construct a point A and a point H (rule W04);

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

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

    % DET: points Hb and A are not the same

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

    % DET: points Hb and H are not the same

  6. Using the point Ha and the line ha, construct a line a (rule W10a);

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

    % NDG: lines a and b are not parallel

    % DET: lines a and b are not the same

  8. Using the line hb and the line a, construct a point B (rule W03);

    % NDG: lines hb and a are not parallel

    % DET: lines hb and a are not the same

animation
Construction in GCLC language

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