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Page 27
... large pair of triangles . Also angles CBG and BCF are equal , since they are similar angles of the smaller pair of triangles . Then subtracting the equal parts from the equal wholes we have equal remainders ( ax . 3 ) , that is the ...
... large pair of triangles . Also angles CBG and BCF are equal , since they are similar angles of the smaller pair of triangles . Then subtracting the equal parts from the equal wholes we have equal remainders ( ax . 3 ) , that is the ...
Page 97
... large parallelogram and ' the parallelo- grams about the diameter ' are each divided by their respec- tive diameters into two triangles , which we know to be equal ( I. 34 ) . Also each large triangle is made up of one of each of the ...
... large parallelogram and ' the parallelo- grams about the diameter ' are each divided by their respec- tive diameters into two triangles , which we know to be equal ( I. 34 ) . Also each large triangle is made up of one of each of the ...
Page 98
... large triangle . D H A K F C G B = From large triangles take away the smaller triangles , and the remainders , the complements , are equal ( ax . 3 ) . ( N.B. This method of proof is very frequently adopted in Book II . ) I. 44. A ...
... large triangle . D H A K F C G B = From large triangles take away the smaller triangles , and the remainders , the complements , are equal ( ax . 3 ) . ( N.B. This method of proof is very frequently adopted in Book II . ) I. 44. A ...
Page 99
... large parallelogram , consequently it must meet the opposite side of the large parallelo- gram which we can obtain by producing FE . But are we sure that HB and FE produced will meet ? We know that any H G B A B E E F straight lines ...
... large parallelogram , consequently it must meet the opposite side of the large parallelo- gram which we can obtain by producing FE . But are we sure that HB and FE produced will meet ? We know that any H G B A B E E F straight lines ...
Page 100
... large parallelogram which we are trying to construct , and which will be completed by drawing a line through K parallel to HF , and producing HA to meet it in L. To obtain the other complement , which is the required parallelogram ...
... large parallelogram which we are trying to construct , and which will be completed by drawing a line through K parallel to HF , and producing HA to meet it in L. To obtain the other complement , which is the required parallelogram ...
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Common terms and phrases
&c.-Q.E.D. PROPOSITION ABCD alternate angles angle ABC angle ACB angles are equal angles equal base BC bisected Cheaper Edition circle cons construction Demy 8vo diameter enunciation equal bases equal in area equal sides equal triangles equilateral triangle Euclid exterior Fcap Frontispiece given angle given line given point given straight line given triangle gnomon gram half the line Illustrations interior and opposite interior angles isosceles triangle Join line equal meet opposite angles opposite sides parallelogram perpendicular Poems problem produced proof Prop rectangle contained rhombus right angles right-angled triangle Second Edition sides equal Small crown 8vo square on AC square on half SUMMARY OF PROPOSITION THEOREM Third Edition Translated triangle ABC triangles are equal twice the rect twice the rectangle vertical angle vols Wherefore whole line
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