This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1878 edition. Excerpt: ...triangle AOB is half the parallelogram BF (I. 41), and the triangle COD is half the parallelogram CF. Therefore in Fig. 1, the triangles A OB and COD are together half the parallelogram ABCD, and in Fig. 2 the difference of the triangles is half the given parallelogram. Again, in Fig. 3, the triangles ...
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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1878 edition. Excerpt: ...triangle AOB is half the parallelogram BF (I. 41), and the triangle COD is half the parallelogram CF. Therefore in Fig. 1, the triangles A OB and COD are together half the parallelogram ABCD, and in Fig. 2 the difference of the triangles is half the given parallelogram. Again, in Fig. 3, the triangles AOB and COD are together o _c _ Fig-3 equal to the triangle BCD. Take away the common triangle COD, and there remains the triangle A OB equal to the triangles BOO and BOD. Therefore the triangle BQD is equal to the difference of the triangles AOB and BOC. Further, in Fig. 4 the triangles AOB and COD are together equal to the triangle BCD. To these equals add the triangle BOC. Then the triangles AOB, BOC and COD are together equal to the figure BOCD. Take away the common triangle COD and there remains the sum of the triangles AOB and BOG equal to the triangle BOD. Q. e. d. 147. Given an indefinite straight line and two points on the same side of it. Find a point in the line such that the straight lines drawn from it to the two given points shall be equally inclined to the given line, and prove that the sum of these straight lines is a minimum. Let CD be the given line, and A, B the two given pointg. Draw AG perpendicular to CD, and produce it until GE equals A G. Join BE, meeting CD in F. F is the required point. For, join AF, and join A and B with any other point H in CD. The angles AFG and GFE are equal (I. 4). Therefore AFG and BFD are equal (I. 15). Also, AF and FB are together equal to EB, and AH and HB are together equal to EH and HB. But EB is less than EH and HB together (I. 20). Therefore the sum of AF and FB is less than the sum of AH and HB. Q. E. D. Cor. Hence, of all triangles on the same base and with a given area the isosceles...
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Add this copy of Exercises on Euclid and in Modern Geometry, New Ed., to cart. $68.07, good condition, Sold by Bonita rated 4.0 out of 5 stars, ships from Santa Clarita, CA, UNITED STATES, published 2016 by Palala Press.
