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Problem 264 from Connelly's corpus: Ea, Hc, Mb


Problem

Given a point Ea, a point Hc and a point Mb, 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 Mb, construct a circle k(Mb,C) (rule W06);

    % NDG: points Hc and Mb are not the same

  3. Using the circle k(Ea,A), the circle k(Mb,C), the point Hc, the point Ea and the point Mb, construct a point A (rule W08);

    % NDG: circles k(Ea,A) and k(Mb,C) intersect

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

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

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

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

    % DET: points Hc and A are not the same

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

    % DET: points Mb and A are not the same

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

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

    % DET: points A and Hb must be different

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

    % DET: points H and Hb are not the same

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

    % NDG: lines c and hb are not parallel

    % DET: lines c and hb are not the same

animation
Construction in GCLC language

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