The Elements of Euclid |
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Page 18
... join FC , GB . a Because AF is equal to AG , and AB to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common b D B A E to the two triangles AFC , AGB ; therefore the base FC is ...
... join FC , GB . a Because AF is equal to AG , and AB to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common b D B A E to the two triangles AFC , AGB ; therefore the base FC is ...
Page 19
... join DC ; there- fore , because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , the two sides DB , BC are equal to the two AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; therefore the ...
... join DC ; there- fore , because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , the two sides DB , BC are equal to the two AC , CB , each to each ; and the angle DBC is equal to the angle ACB ; therefore the ...
Page 21
... join DE and upon it describe b an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAC . sides EA , AF , Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; the two sides DA , AF ...
... join DE and upon it describe b an equilateral triangle DEF ; then join AF ; the straight line AF bisects the angle BAC . sides EA , AF , Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; the two sides DA , AF ...
Page 22
... join FC ; the straight line FC drawn . from the given point C is at right angles to the given straight line AB . A D E B Because DC is equal to CE , and FC common to the two triangles DCF , ECF ; the two sides DC , CF , are equal to the ...
... join FC ; the straight line FC drawn . from the given point C is at right angles to the given straight line AB . A D E B Because DC is equal to CE , and FC common to the two triangles DCF , ECF ; the two sides DC , CF , are equal to the ...
Page 26
... join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB are equal to CE , EF , each to each ; and the angle AEB is equal to the angle CEF , because ...
... join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB are equal to CE , EF , each to each ; and the angle AEB is equal to the angle CEF , because ...
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Common terms and phrases
altitude angle ABC angle BAC base BC BC is equal BC is given bisected Book XI centre circle ABCD circumference cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gnomon greater join less Let ABC meet multiple parallel parallelogram parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR third triangle ABC triplicate ratio vertex wherefore