Elementary Geometry Volume 2 (Paperback)

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. 1868 edition. Excerpt: ... about each of the angles of two triangles are proportionals, the triangles will be similar. Let ABC, DEF be two triangles which have their sides A. about each of their angles proportional, that is, AB: BC:: DE: EF, and BC: CA-: EF: FD, and CA: AB:: FD: DE. Conceive a triangle equiangular to ABC applied to EF, on the opposite side of the base EF, so that the angles PEG, EFG are equal to B and C respectively. Then the triangle GEF is equiangular to ABC, and therefore similar to it, and therefore GE: EF:: AB: BC, hut AB: BC:: DE: EF, and therefore GE: EF:: DE: EF, and therefore GE = ED. Similarly GF= DF, aad the triangle DEF is therefore equiangular to GEF, and therefore also to ABC. Therefore the triangle DEF is similar to the triangle ABC. This theorem is a generalization of Book 1. Theorem 18. If the three sides of one triangle are-respectively equal to the three sides of another, these triangles will be equal in all respects. Theorem 9. If two triangles have the sides about an angle of the one triangle proportional to the sides about an angle of the other, and have also the angle opposite that which is not the less of the two sides of the one equal to the corresponding angle of the other, these triangles will be similar. Let ABC, DEF be the two triangles, in which BA: AC:: ED: DF, and let AC be not less than AB, and DF therefore not less than DE, and also let the angle B = the angle E. Then shall the triangles be similar. Cut off BD" = ED, and draw to BC an oblique DT parallel to A C. Then by similar triangles BDF and BA C, BD: DF' BA: AC, that is, ED: DF:: BA: A C, but ED: DF:: BA: AC, therefore DF= DF, and since DF is not less than DE, the two triangles BDF', EDF are equal in all respects by Book 1. Theorem 19, that is, the triangle EDF...