Geometrical Researches On the Theory of Parallels

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If, finally, the point H falls outside the triangle c ABC (Fig. 18), the perpendicular CG- goes, in Fig. 17. consequence, through the triangle, and so we go over from the triangle ABC to the quadrilateral AFGC by adding the triangle FAD = DBH, and then taking away the triangle CGE — EBH.
Supposing in the spherical quadrilateral AFGC a great circle passed through the points A and G, as also through F and C, then will their arcs between AG and FC equal one another (Theorem 15), consequently also
...the triangles FAC and ACG be congruent (Theorem 15), and the angle FAC equal the angle ACG.
Hence follows, that in all the preceding cases, the sum of all three angles of the spherical triangle equals the sum of the two equal angles in the quadrilateral which are not the right angles.
Therefore we can, for every spherical triangle, in which the sum of the three angles is S, find a quadrilateral with equivalent surface, in which are two right angles and two equal perpendicular sides, and where the two other angles are each £S.


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