Artículos relacionados a Euclid for beginners, books I and II with exercises
Sinopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ...(ax. 1) The parallelogram ABCD=the parallelogram EFGH. Wherefore, Parallelograms on equal bases, &c. Q. E. D. Prop. XXXVII. Theorem. Triangles on the same base and between the same parallels are equal to one another. Let ABC and DEC be triangles on the same base BC, and between the same parallels AD and BC. Then it is to be proved that The triangle ABC = the triangle DISC. F Construction.--1. Produce AD both ways to points E and F. 2. Through B draw BE parallel to CA, and through C draw CF parallel to BD (I. 31). Proof.--Because each of the figures EBCA and DBCF is a parallelogram (def. 34), and because they are on the same base BC, and between the same parallels BC and EF, therefore the parallelogram EBCA = the parallelogram DBCF (I. 35). Next, because the triangle ABC is half the parallelogram EBCA, and the triangle DEC is half the parallelogram DBCF (I. 34), Therefore, it is proved, as required, that The triangle ABC = the triangle DBG (ax. 7).' Wherefore, Triangles on the same base, &c, Q. E. D. Prop. XXXVIII. Theorem. Triangles on equal bases and between the same parallels are equal to one another. Let ABC and DEF be triangles on equal bases BC and EF, and between tlie same parallels AD and BF. Then it is to be proved that The triangle ABC = the triangle DEF. Construction.--1. Produce AD both ways to G and H. 2. Through B draw BG parallel to CA, and through F draw FH parallel to ED (I. 31). Proof.--Because each of the figures GBCA and DEFH is a parallelogram (def. 34), and because they are on equal bases BC and EF, and between the same parallels GH and BF, therefore the parallelogram GBCA = the parallelogram DEFH (I. 36). Next, because the triangle ABC is half the parallelogram GBCA, and because the triangle DEF is half the parallelogram DEFH (...
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Reseña del editor
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ...(ax. 1) The parallelogram ABCD=the parallelogram EFGH. Wherefore, Parallelograms on equal bases, &c. Q. E. D. Prop. XXXVII. Theorem. Triangles on the same base and between the same parallels are equal to one another. Let ABC and DEC be triangles on the same base BC, and between the same parallels AD and BC. Then it is to be proved that The triangle ABC = the triangle DISC. F Construction.--1. Produce AD both ways to points E and F. 2. Through B draw BE parallel to CA, and through C draw CF parallel to BD (I. 31). Proof.--Because each of the figures EBCA and DBCF is a parallelogram (def. 34), and because they are on the same base BC, and between the same parallels BC and EF, therefore the parallelogram EBCA = the parallelogram DBCF (I. 35). Next, because the triangle ABC is half the parallelogram EBCA, and the triangle DEC is half the parallelogram DBCF (I. 34), Therefore, it is proved, as required, that The triangle ABC = the triangle DBG (ax. 7).' Wherefore, Triangles on the same base, &c, Q. E. D. Prop. XXXVIII. Theorem. Triangles on equal bases and between the same parallels are equal to one another. Let ABC and DEF be triangles on equal bases BC and EF, and between tlie same parallels AD and BF. Then it is to be proved that The triangle ABC = the triangle DEF. Construction.--1. Produce AD both ways to G and H. 2. Through B draw BG parallel to CA, and through F draw FH parallel to ED (I. 31). Proof.--Because each of the figures GBCA and DEFH is a parallelogram (def. 34), and because they are on equal bases BC and EF, and between the same parallels GH and BF, therefore the parallelogram GBCA = the parallelogram DEFH (I. 36). Next, because the triangle ABC is half the parallelogram GBCA, and because the triangle DEF is half the parallelogram DEFH (...
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