| Edward Olney - Geometry - 1883 - 333 pages
...falls between CA and CB, and CH will meet AB in some point, as H. Draw HE. The triangles HCB and HCE **have two sides and the included angle of the one, equal to** the corresponding parts of the other, whence HE = HB (?). Now AH + HE > AE but AH + HE = AH + HB =... | |
| Geometry - 1884 - 155 pages
...similar. Take AG equal to DE, and AH to DF; also, join GH. Then the triangles AGH, DEF, having two **sides and the included angle of the one equal to two sides and the included angle** of the other, are equal throughout (Theo. XII, Book I). Now, by hypothesis, Hence, it follows that... | |
| Geometry - 1884 - 155 pages
...DB < AC + CB. NOTE. — Have pupils give demonstration, prolonging BD instead of AD. THEOREM XII. If **two triangles have two sides and the included angle of the one,** respectively, equal to two sides and the included, angle of the other, they are equal throughout. The... | |
| Charles Davies - Geometry - 1885 - 512 pages
...CAG equal to the angle D (Post. 7) ; make AG equal to DE, and draw GC. Then the triangles AGC and DEF **have two sides and the included angle of the one equal...and the included angle of the other, each to each** ; consequently, GC is equal to EF (PV). Now, the point G may be without the triangle ABC, it may be... | |
| James Wallace MacDonald - Geometry - 1889 - 63 pages
...Proposition XLV. A Theorem. 87. Two parallelograms are equal if they have two sides and the included angle of **one equal to two sides and the included angle of the other, each to each.** Proposition XLVI. A Theorem. VIII. POLYGONS OF MORE THAN FOUR SIDES. 89. What is a polygon ? a. A pentagon?... | |
| James Wallace MacDonald - Geometry - 1894 - 65 pages
...Proposition XLV. A Theorem. 87. Two parallelograms are equal if they have two sides and the included angle of **one equal to two sides and the included angle of the other, each to each.** Proposition XLVI. A Theorem. 88. Parallel lines are everywhere equally distant. VIII. POLYGONS OF MORE... | |
| Randal Charles John Nixon - Geometry - 1890 - 400 pages
...BC = side YZ, side CA = side ZX ; and, finally, area ABC = area XYZ. Proposition 4. THEOREM — If **two triangles have two sides and the included angle of the one,** respectively equal to two sides and the included angle of the other, then the triangles are identically... | |
| Seth Thayer Stewart - Geometry - 1891 - 406 pages
...equal to the given lines, and it is, therefore, the required triangle. PROPOSITION I. Theorem : // **two triangles have two sides and the included angle...and the included angle of the other, each to each,** they are equal in all their parts. Statement : Let the two triangles 123 and 456 have the side / 2... | |
| Seth Thayer Stewart - Geometry - 1891 - 406 pages
...coincide, and are equal. Conclusion \ABCK and E FS H being, etc. Corollary I. Parallelograms having two **sides and the included angle of the one equal to two...and the included angle of the other, each to each,** are equal. PROPOSITION XVIII. 21O. Theorem : A parallelogram is divided by either of its diameters... | |
| Thomas J. Foster - Coal mines and mining - 1891 - 415 pages
...opposite or vertical angles are equal. 4. If two angles have their sides parallel they are equal. 5. If **two triangles have two sides and the included angle...the one, equal to two sides and the included angle** of the other, they are equal in all their parts. If two triangles have two angles and the included... | |
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