By P. L. Garcia, A. Perez-Rendon, J. M. Souriau

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Construct points G2 , G3 , . . , Gn along AB with AG1 ∼ = G 1 G2 ∼ = ··· ∼ = Gn−1 Gn . ← → Let G2 K be the perpendicular to AC from G2 at K. We claim that K = H2 . Let AL be perpendicular to AC with AL ∼ = G1 H1 . Join LG1 . Since angles ∠AG1 H1 and ∠G1 AL are complementary to ∠G1 AH1 , they are congruent. ) By Side–Angle–Side, G1 AH1 and AG1 L are congruent and so AH1 ∼ = LG1 and ∠ALG1 is a right angle. Let M on G2 K be such that KM ∼ = H1 G1 . By the same argument ∠KMG1 is congruent to a right angle and G1 M ∼ = H1 K.

Triangles on the same base and of equal areas are on the same parallels. 40. Triangles on congruent bases on the same side of a given line and of equal areas are also in the same parallels. 41. If a parallelogram has the same base with a triangle and they are in the same parallels, the parallelogram has twice the area of the triangle. 42. In a given angle, one can construct a parallelogram of area equal to the area of a given triangle. 43. In any parallelogram, the complements of the parallelogram about the diameter are equal in area.

In order to make the definition of right angle and Postulate IV precise, we need a notion of congruence of angles. Congruence of figures may be thought of as a motion in space of one figure that superimposes it on another figure such that corresponding points and line segments coincide. 26. Bertrand Russell (1872–1970), in his 1902 article for the Encyclopedia Britannica Supplement, points out that “(a)ctual superposition . . ” We take a congruence, a “transference of attention,” to mean a mapping φ : S → S, defined on the underlying set S of an incidence geometry and satisfying: (1) φ is a one-to-one correspondence.