The elements of Euclid for the use of schools and colleges : comprising the first two books and portions of the eleventh and twelfth books; with notes and exercises . ual to the triangleGEF, [Axiom 1. the greater to the less: which is AG is, not parallel to BF. In the same manner it can be shewn that no otherstraight line through A but AB is parallel to BF jtherefore AD is parallel to BF. Wherefore, equal triangles &c. PROPOSITION 41. THEOREM. If a parallelogram and a triangle he on the same baseand between the same parallels, the parallelogram sliall heduiible of t
The elements of Euclid for the use of schools and colleges : comprising the first two books and portions of the eleventh and twelfth books; with notes and exercises . ual to the triangleGEF, [Axiom 1. the greater to the less: which is AG is, not parallel to BF. In the same manner it can be shewn that no otherstraight line through A but AB is parallel to BF jtherefore AD is parallel to BF. Wherefore, equal triangles &c. PROPOSITION 41. THEOREM. If a parallelogram and a triangle he on the same baseand between the same parallels, the parallelogram sliall heduiible of the triangle. Let the parallelogram ABCD and the triangle EEC the same base BC, and between the same parallelsliC. AE : the parallelogram ABCD shall be double of thetriangle EBC. Join AC. Tlicn - the triangle ABCis equal to the triangle EBC,because they are on the samebase BC, and between the sameparallels ^(7,^£. [ Jjiit the parallelogram ABCD ,is double of the triangle ABC,because the diameter yiC bisects the parallelogram. [J. the parallelogram ABCD is also double of thetriangle EBC. \i\iQVQioTQ, if a parallelogram &c. 44 EUCLIDS ELEMENTS. PROPOSITION 42. PROBLEM, To describe a parallelogram that shall he equal to agiven triangle, and have one of its angles equal to a givenrectilineal angle. Let ABC be the given triangle, and D the given recti-lineal angle : it is required to describe a parallelogram thatshall be equal to the given triangle ABC, and have one ofits angles equal to Z). Bisecti?aat^:[ AE, and at the pointE, in the straight line EC,make the angle C^i^equaltoi>; [ through A draw AFGparallel to EC, and throughC draw CG parallel toEF. [I. 31. Therefore FECG is a parallelogi-am. [Definition. And, because BE is equal to EC, [Construction* the triangle ABE is equal to the triangle AEC, becausethey are on equal bases BE, EC, and between the sameparallels BC, AG. [I. 38. Therefore the triangle ABC is double of the triangl
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