docs: boolean leftOf(double[] a, double[] b, double[] d) {
      final double bax = b[0] - a[0], bay = b[1] - a[1];
      final double dax = d[0] - a[0], day = d[1] - a[1];
      final double cross = bax * day - bay * dax;
      return cross > 1e-10 * Math.max(Math.max(bax, bay), Math.max(dax, day));
    }