 | John Keill - Logarithms - 1723 - 444 pages
...Angle KCF; and the Right Angle FHC equal to the Right Angle FKC; the two Triangles FHC } FKC fhall have two Angles of the one equal to two Angles of the other, and one Side of the one equal to one Side of the other, viz. the Side FC common to each of them. And fo... | |
 | Euclid, John Keill - Geometry - 1733 - 444 pages
...F. But the Right Angle AFE is equal to the Right Angle BFE ; therefore the two Triangles EAF, EBF, have two Angles of the one equal to two Angles of the other, and the Side EF is common to both. Wherefore the other Sides 146. i. of the one fhall be f equal to the... | |
 | Robert Simson - Trigonometry - 1762 - 488 pages
...bifttfed by BD, and that the right angle BED is equal to the right angle BFD, the two ttiangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is oppoilte to one of the equal angles in each, is com-rv •*• G C nwn to both... | |
 | John Keill - Geometry - 1772 - 462 pages
...Angle KCF, and the Right An»ie FHC equal to the Right Angle FKC; the twoTiiangles FHC, FKC, fhall have two Angles of the one equal to two Angles of the other, and one Side of the one equal to one Side of the other, viz. the SideF C common to each of them : the Angle... | |
 | Robert Simson - Trigonometry - 1781 - 534 pages
...bifected by BD, and that the right angle BED is equal to the right angle BFD, the two triangles EBD,F6D have two angles of the one equal to two angles of the other, and the fide BD, which is oppofite to one of the TJ T-» ^ e"qual angles in each, is common ** ,*- ' to... | |
 | John Keill - Geometry - 1782 - 476 pages
...Angle FBD, and the Right Angle BED is equal to the Rifht Angh BFD ; then the two Triangles EBD, DBF) have two Angles of the one, equal to two Angles of the other, and one Side DB common to both, viz. viz. that which fubttnds the equal Angles';' therefore the other Sides... | |
 | Euclid, John Playfair - Euclid's Elements - 1795 - 462 pages
...KGN ; therefore the bafe KL is equal to the bafe KN<1. But becaufe the triangle KGN is ifof- a celes, the angle GKN is equal to the angle GNK, and the angles GMK, GMN are both right angles by conírruítíon; wherefore, the triangles GMK, GMN Ime two angles of the one equal to two angles of... | |
 | Alexander Ingram - Trigonometry - 1799 - 374 pages
...biftcted by BD, and that the right angle BED Is equal to the righr angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is oppciite to one of the equal angles in each, is common to C both ; therefore... | |
 | Tiberius Cavallo - Aeronautics - 1803 - 638 pages
...die angle FGD is equal to the angle CGD; whence it follows, that the triangles DGC and DGF, Tiaving two angles of the one equal to two angles of the other, and a correfpondent fide, viz. DG, common, are equal in every refpect J ; * It is ufelefs to take notice... | |
 | John Playfair, Euclid - Circle-squaring - 1804 - 468 pages
...therefore the bafe KL is equal to the bafe KN <i. But becaufe the triangle KGN is ifof- d + .i. celes, trie angle GKN is equal to the angle GNK and the angles GMK, GMN are both right angles by conftruchoa; wherefore, the triangles GMK, GMN have two angles of the one equal to two nugles of the... | |
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But because the triangle KGN is isosceles, the angle GKN is equal to the angle GNK, and the angles GMK, GMN are both right angles by construction ; wherefore, the triangles GMK, GMN have two angles of the one equal to two angles of the other, and they... 
