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" SIMILAR triangles are to one another in the duplicate ratio of their homologous sides. "
The school edition. Euclid's Elements of geometry, the first six books, by R ... - Pagina 271
door Euclides - 1864
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Modern geometry [ed.] with an appendix by W.B. Jack

Richard Wormell - 1876 - 268 pagina’s
...circumscribed circles, or their proportionals the radii of the inscribed circles. THEOREM LXXXV. (¿.) Similar triangles are to one another in the duplicate ratio of their homologous sides. Let А В С, DEF be similar triangles having Z. B = ¿EandAB:DE = BC:EFso that В С and E^F are homologous...
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Syllabus of Plane Geometry: (corresponding to Euclid, Books I-VI) ...

Association for the improvement of geometrical teaching - 1876 - 66 pagina’s
...to one another the ratio compounded of the ratios of their bases and of their altitudes. THEOR. 15. Similar triangles are to one another in the duplicate ratio of their homologous sides. THEOR. 16. The areas of similar rectilineal figures are to one another in the duplicate ratio of their...
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Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...

Robert Potts - 1876 - 446 pagina’s
...equal to one angle of the other, have their sides about the equal angles reciprocally proportional. 4. Similar triangles are to one another in the duplicate ratio of their homologous sides. 5. In any right angled triangle, any rectilineal figure described on the side subtending the right...
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Elements of geometry, containing books i. to vi.and portions of books xi ...

Euclides, James Hamblin Smith - 1876 - 376 pagina’s
...described, on a given line, similar to a given fig. QEF PROPOSITION XIX. THEOREM. Similar triangles an to one another in the duplicate ratio of their homologous sides. Let ABC, DBF be similar A s, having / s at A, B, C= L s at D, E, F respectively, so that BC and EF are homologous...
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Dublin examination papers

Dublin city, univ - 1876 - 420 pagina’s
...the other segment. If the whole line be 10 feet long, find the lengths of the segments. 2. Prove that similar triangles are to one another in the duplicate ratio of their sides. Divide a triangle into three equal parts by right lines drawn parallel to its base. 3. Divide...
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Mathematical Exercises ...: Examples in Pure Mathematics, Statics, Dynamics ...

Samuel H. Winter - 1877 - 452 pagina’s
...have the same base, are to one another as their altitudes. 7. Define duplicate ratio, and prove that similar triangles are to one another in the duplicate ratio of their homologous sides. On the side AB of a triangle ABC, AD is taken equal to one third of AB ; and on AC, AE is taken equal...
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Solutions to the mathematical examination papers set for admission to the ...

D. Tierney - 1877 - 126 pagina’s
...HE, HF. Then EHF is the triangle required, for it is isosceles and equal to EFG, that is, to ABC. 10. Similar triangles are to one another in the duplicate ratio of their homologous sides. 11. Given a point O in the line AB, find two other points 0, J?, such that a line OP given in direction...
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Elementary Geometry ...

James Maurice Wilson - 1878 - 450 pagina’s
...triangle DEF in the duplicate ratio of BC to EF. THEOREM 16. The areas of similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. Let ABCDE, PQ_RST\3z similar polygons. JD S Divide each of them into the same number of similar triangles...
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The elements of plane geometry, from the Sansk. text of Ayra Bhatta, ed. by ...

Āryabhaṭa - 1878 - 100 pagina’s
...and this has been proced of triangles (P. 34). Therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. Cor. 2. If to AB and FG, two of the homologous sides ; of the polygon, a third proportional M Is taken...
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The Elements of Plane and Solid Geometry: With Chapters on Mensuration and ...

Isaac Sharpless - 1879 - 282 pagina’s
...resulting polygons similar. Proposition 19. Theorem.—Similar triangles are to one another as the squares of their homologous sides. Let ABC, DEF be similar...having the angle B equal to the angle E, and let AB : BC :: DE : EF, so that BC and EF are homologous sides; then ABC : DEF ::BC* : EF\ Because ABC, DEF...
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