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Problem 420 from Wernick's corpus: Mb, Hb, Hc


Problem

Given a point Mb, 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 Mb and the point Hb, construct a line b (rule W02);

    % DET: points Mb and Hb 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(Mb,C) and the line b, construct a point C and a point A (rule W04);

    % NDG: line b and circle k(Mb,C) intersect

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

    % DET: points Hc and A are not the same

  5. Using the point Hb and the line b, construct a line hb (rule W10b);

  6. 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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