| Elias Loomis - Conic sections - 1849 - 226 pages
...CD, and BC is common to the two triangles ABC, BCD, the two triangles ABC, BCD have two sides and C **the included angle of the one, equal to two sides and the included angle** of the other; therefore, the side AC is equal to BD (Prop. VI.), and the angle ACB to the angle CBD.... | |
| CHARLES DAVIS, L.L. D. - 1850
...angle BDC. Of Triangles. ACD+BCD=ADC+BDC that is, the angle ACB=ADB. Now, the two triangles ACB and ADB **have two sides and the included angle of the one equal to two sides and the** ineluded angle of the other, each to each: hence, the remaining angles will be equal (Th. iv) : consequently,... | |
| Charles Davies - Geometry - 1850 - 216 pages
...angle BDC. Of Triangles. ACD+BCD=ADC+BDC that is, the angle ACB=ADB. Now, the two triangles ACS and ADB **have two sides and the included angle of the one equal to two sides and the** in. eluded angle of the other, each to each : hence, the remaining angles will be equal (Th. iv) :... | |
| George Roberts Perkins - 1850
...in the same manner it may be shown that the angle AFD is equal to BFC. PROPOSITION III. THEOREM. If **two triangles have two sides and the included angle of the one, equal to** the two sides and the included angle of the other, the triangles will be identical, or equal in all... | |
| Charles Davies - Geometry - 1886 - 324 pages
...we have Of Triangles. ACD+BCD=ADC+BDC that is, the angle ACB=ADB. Now, the two triangles ACB and ADB **have two sides and the included angle of the one equal to two sides and the** in. eluded angle of the other, each to each: hence, the remaining angles will be equal (Th. iv) : consequently,... | |
| A. M. LEGENDRE - 1852
...have, AB : DE :: AC : and by construction, AG=DE: hence, AH=DF. Therefore, the two triangles AGH, DEF, **have two sides and the included angle of the one equal to** the sides and the included angle of the other : hence, they are equal (B. L; p. 5) ; but the triangle... | |
| Charles Davies, Adrien Marie Legendre - Geometry - 1854 - 432 pages
...: DE : : AC : DF; and by construction, AG=DE : hence AN=DF. Therefore, the two triangles A GH, DEF, **have two sides and the included angle of the one equal to** twd sides and the included angle of the other: hence, they are equal (BI, p. 5); but the triangle AGH... | |
| S.S. SCHMUCKER D.D. - 1855
...the nature, eg, of the demonstration of the theorem of Euclid known as Prop. IV. of Book I., viz.: If **two triangles have two sides, and the included angle...the one equal to two sides and the included angle** of the other, they must be identical or equal in all respects. For it virtually consists in supposing... | |
| GEORGE R. PERKINS - 1856
...line which has the sum of the interior angles less than two right angles. OF TRIANGLES. THEOREM XX. If **two triangles have two sides and the included angle of the one, equal to** the two sides and the included angle of the other, the triangles will fie identical, or equal in all... | |
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