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" BAC equal to the third angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater, and make BH equal to EF, [I. "
Elements of Geometry, Conic Sections, and Plane Trigonometry - Page 31
by Elias Loomis - 1880
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Euclid's Elements of geometry [book 1-6, 11,12] with explanatory notes ...

Euclides - 1845 - 546 pages
...be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDF. For if BC be not equal to EF, one of them must be greater than the other. Let BC be greater than EF; make BH equal to EF, (i.3.) and join AH. Then in the two triangles ABH, DEF, because...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1849 - 252 pages
...side of the one, equal to the hypothenuse and a side of the other, each to each, the triangles are equal. Let ABC, DEF be two right-angled triangles,...them being right angles, the two triangles are equal (Prop. VI.), and AG is equal to DF. But, by hypothesis, AC is equal to DF, and therefore AG is equal...
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The Elements of geometry; or, The first six books, with the eleventh and ...

Euclides - 1855 - 262 pages
...to DF, and B С to E F. And the third angle BA С is equal to the third angle EDF. For, if B С be not equal to EF, one of them must be greater than the other. Let B С be the greater of the two. Make BH equal (I. 3) to EF, and join A H. Because in the two triangles...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1857 - 242 pages
...of the other; then will & CE V the side BC be equal to EF, and the triangle ABC to the triangle DBF. For if BC is not equal to EF, one of them must be...the less, and join AG. Then, in the triangles ABG, DBF, because AB is equal to DE, BG is equal to EF, and the angle B equal to the angle E, both of them...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1858 - 256 pages
...equal to the hypothenuse and a side of the other each to each, the triangles are equal. Let ABC, DBF be two right-angled triangles, having the hypothenuse...DEF, because AB is equal to DE, BG is equal to EF, anu the angle B equal to the angle E, both of them being right angles, the two triangles are equal...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1860 - 246 pages
...of the other; then will the side BC be equal to EF, and the triangle ABC to the triangle DEF. & CE For if BC is not equal to EF, one of them must be...DEF, because AB is equal to DE, BG is equal to EF, ana the angle B equal to the angle E, both of them being right angles, the two triangles are equal...
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Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...

Robert Potts - Geometry, Plane - 1860 - 380 pages
...equal, AC to DF and BC to EF, and also the third angle BA C to the third angle EDF AD BHC For if BC be not equal to EF, one of them must be greater than the other. If possible, let BC be greater than EF; make BH equal to EF, (l. 3.) and join AH. Then in the two triangles...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1861 - 254 pages
...a side of the one, equal to the hypothenuse and a side of the other each to each, the triangles are equal. Let ABC, DEF be two right-angled triangles,...DEF, because AB is equal to DE, BG is equal to EF, ana the angle B equal to the angle E, both of them being right angles, the two triangles are equal...
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Elements of Geometry and Conic Sections

Elias Loomis - Conic sections - 1861 - 244 pages
...the side BC be equal to EF, and the triangle ABC to the triangle DEF. For if BC is not equal to EF9 one of them must be greater than the other. Let BC...them being right angles, the two triangles are equal (Prop. VI.), and AG is equal to DF, But, by hypothesis, AC is equal to DF, and therefore AG is equal...
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Euclid's Elements of geometry, books i. ii. iii. iv

Euclides - 1862 - 140 pages
...AC to DF, and BC to EF; and also the angle BAC to the angle EDF. Hypothesis.— (II.) For if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two. Construction. — Make BH equal to EF (I. 3), and join AH. EF (const.), and AB is equal...
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