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algebraically altitude axis base base ABC cause circle circle ABCD circumference circumscribed compounded cone consequently construction contained COROLLARY cylinder definition DEMONSTRATION described diameter divided double draw drawn EFGH equal angles equiangular equilateral equimultiples expressed follows fore four fourth greater greater ratio half inscribed join less likewise magnitudes manner meet multiple opposite parallel parallelogram passing perpendicular plane polygon prism PROBLEM produced proportional PROPOSITION proved pyramid pyramid ABCG ratio reason rectangle rectilineal figure remaining right angles SCHOLIUM segments shown sides similar similarly solid angle solid CD solid parallelopipeds SOLUTION sphere square straight line taken THEOREM THEOREM.-If third triangle ABC triplicate ratio vertex wherefore whole
Page 198 - ... have an angle of the one equal to an angle of the other, and the sides about those angles reciprocally proportional, are equal to une another.
Page 115 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Page 82 - From the point A draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.
Page 198 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Page 53 - Convertendo, by conversion ; when there are four proportionals, and it is inferred, that the first is to its excess above the second, as the third to its excess above the fourth.
Page 40 - A and B are not unequal ; that is, they are equal. Next, let C have the same ratio to each of the magnitudes A and B ; then A shall be equal to B.
Page 119 - For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK.
Page 115 - FB ; (i. 4.) for the same reason, CF is equal to FD : and because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each ; and the base DF was proved equal to the base FC ; therefore the angle FAD is equal to the angle FBC: (i. 8.) again, it was proved that GA is equal to BH, and also AF to FB; therefore FA and AG are equal...
Page 94 - C, they are equiangular, and also have their sides about the equal angles proportionals (def. 1. 6.). Again, because B is similar to C, they are equiangular, and have their sides about the equal angles proportionals (def. 1. 6.) : therefore the figures A, B, are each of them equiangular to C, and have the sides about the equal angles of each of them, and of C, proportionals. Wherefore the rectilineal figures A and B are equiangular (1. Ax. 1.), and have their sides about the equal angles proportionals...