Class lessons on Euclid1882 |
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Side 16
... gram parallelogram A triangle sq . square .. therefore ..since or because PROP . I. Having now cleared the way , we can enter upon the discussion of the first proposition , which is a problem , that is , it shows how something is to be ...
... gram parallelogram A triangle sq . square .. therefore ..since or because PROP . I. Having now cleared the way , we can enter upon the discussion of the first proposition , which is a problem , that is , it shows how something is to be ...
Side 86
... gram are equal . II . △ ABC = BCD in area , .. diameter bisects || gram ABDC . III . / CAB = / BDC ; △ ACD ( ACB + BCD ) = 2 ABD ( ABC + CBD ) .. opposite angles of gram are equal . Let us briefly recapitulate what we have learnt ...
... gram are equal . II . △ ABC = BCD in area , .. diameter bisects || gram ABDC . III . / CAB = / BDC ; △ ACD ( ACB + BCD ) = 2 ABD ( ABC + CBD ) .. opposite angles of gram are equal . Let us briefly recapitulate what we have learnt ...
Side 89
... gram EFCB = FDC + gram ADCB . Therefore if we can show that the triangles are equal , the remainders , the parallelograms , must be equal ( ax . 3 ) . Let us examine these triangles : = : - side BE FC since they are opposite side of gram ...
... gram EFCB = FDC + gram ADCB . Therefore if we can show that the triangles are equal , the remainders , the parallelograms , must be equal ( ax . 3 ) . Let us examine these triangles : = : - side BE FC since they are opposite side of gram ...
Side 90
... gram is double of △ DBC ( I. 34 ) . ..grams are equal . Cases II . and III . A AEBA DFC ( I. 4 , I. 8 , I. 26 ) . △ DFC + || gram ADCB = ^ AEB + || gram EFCB . .. || gram ADCB = ram EFCB ( ax . 3 ) . ( N.B. Equal parts of the ...
... gram is double of △ DBC ( I. 34 ) . ..grams are equal . Cases II . and III . A AEBA DFC ( I. 4 , I. 8 , I. 26 ) . △ DFC + || gram ADCB = ^ AEB + || gram EFCB . .. || gram ADCB = ram EFCB ( ax . 3 ) . ( N.B. Equal parts of the ...
Side 91
... gram ADCB = || gram EFGH . I. 37. In the next two theorems we prove the corresponding cases of equality of area in triangles . Proposition XXXVII . shows that triangles on the same base and between the same parallels are equal in area ...
... gram ADCB = || gram EFGH . I. 37. In the next two theorems we prove the corresponding cases of equality of area in triangles . Proposition XXXVII . shows that triangles on the same base and between the same parallels are equal in area ...
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Vanlige uttrykk og setninger
&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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