If two triangles have two sides, and the included angle of the one equal to two sides and the included angle of the other, each to each, the two triangles are equal in all respects. Elements of Plane Geometry - Page 48by Thomas Hunter - 1878 - 132 pagesFull view - About this book
| Charles Davies - Logic - 1850 - 402 pages
...to each, and the triangles themselves will be equal." Prop. V. " When two triangles have two tides **and the included angle of the one, equal to two sides...and the included angle of the other, each to each,** the two triangles will be equal." Axiom I. " Things which are equal to the same thing, are equal to... | |
| Charles Davies - Geometry - 1850 - 236 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) :... | |
| Charles Davies - Geometry - 1850 - 218 pages
...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,... | |
| George Roberts Perkins - Geometry - 1850 - 332 pages
...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 - 340 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,... | |
| Adrien Marie Legendre - Geometry - 1852 - 436 pages
...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... | |
| Anthony Dumond Stanley - Sphere - 1854 - 130 pages
...these two sides. Now DCE is equal to this triangle, two sides and the included angle of the one being **equal to two sides and the included angle of the other, each to each** ; since CE and DE are supplements of AE and BE, and the angle CED is equal to AEB. Since then CDE is... | |
| Charles Davies - Geometry - 1854 - 436 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... | |
| Charles Davies, William Guy Peck - Electronic book - 1855 - 592 pages
...angles will also be equal, each to each, and the triangles themselves will bo equal." Prop. V. " When **two triangles have two sides, and the included angle...and the included angle of the other, each to each,** the two triangles will be equal. " Axiom 1. Things which are equal to the same thing are equal to each... | |
| Charles Davies - Geometry - 1855 - 336 pages
...parallel to DC, the alternate angles, ABD and BDC are equal (Th- xii) : moreover, the side BD is common ; **hence the two triangles have two sides and the included...the one, equal to two sides and the included angle** of the other: the triangles are therefore equal, and consequently, AD is equal to BC, and the angle... | |
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