Geometrical Exercises for BeginnersMacmillan, 1882 - 203 sider |
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Side 2
... parallels ( I. 39 ) . Hence FE is parallel to the side BC . Similarly the line ED is parallel to the side AB ; therefore BFED is a parallelo- gram , and therefore ( I. 34 ) FE , BD are equal : that is , FE is equal to half the base ...
... parallels ( I. 39 ) . Hence FE is parallel to the side BC . Similarly the line ED is parallel to the side AB ; therefore BFED is a parallelo- gram , and therefore ( I. 34 ) FE , BD are equal : that is , FE is equal to half the base ...
Side 1
... parallel to the third side and equal to half of it . A F E B D Let D , E , F , be the middle points of the sides BC , CA , AB , of any triangle ABC . Join FE , BE , CF , and ED . C. G. 1 As the triangles AFE , BFE stand upon equal bases ...
... parallel to the third side and equal to half of it . A F E B D Let D , E , F , be the middle points of the sides BC , CA , AB , of any triangle ABC . Join FE , BE , CF , and ED . C. G. 1 As the triangles AFE , BFE stand upon equal bases ...
Side 2
... parallels ( I. 39 ) . Hence FE is parallel to the side BC . Similarly the line ED is parallel to the side AB ; therefore BFED is a parallelo- gram , and therefore ( I. 34 ) FE , BD are equal : that is , FE is equal to half the base ...
... parallels ( I. 39 ) . Hence FE is parallel to the side BC . Similarly the line ED is parallel to the side AB ; therefore BFED is a parallelo- gram , and therefore ( I. 34 ) FE , BD are equal : that is , FE is equal to half the base ...
Side 3
... parallel to BC ( I. ) ; therefore ( I. 29 ) the angles AEO , A CB are equal ; but ACB is a right angle , therefore ... parallel lines be drawn meeting a third indefinite straight line : ( 1 ) the sum of the two external parallels is ...
... parallel to BC ( I. ) ; therefore ( I. 29 ) the angles AEO , A CB are equal ; but ACB is a right angle , therefore ... parallel lines be drawn meeting a third indefinite straight line : ( 1 ) the sum of the two external parallels is ...
Side 4
... parallel to LMN , and meeting PL , OM , produced if necessary , in R , T respectively . Since O is the middle point of PQ , and OT is parallel to the base PR of the triangle PQR ; therefore ( 1 , Note ) 20T = PR . And since the opposite ...
... parallel to LMN , and meeting PL , OM , produced if necessary , in R , T respectively . Since O is the middle point of PQ , and OT is parallel to the base PR of the triangle PQR ; therefore ( 1 , Note ) 20T = PR . And since the opposite ...
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Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Geometrical Exercises: For Beginners (1882) Samuel Constable Ingen forhåndsvisning tilgjengelig - 2008 |
Vanlige uttrykk og setninger
arranged Assistant base base angles Bisect BOOK Cambridge centre chord circle circumference clearly cloth common Consider constant construct containing Crown 8vo cutting describe diameter difference divide double draw drawn Edition English enlarged equal equal to half Examples EXERCISES external bisector Extra fcap Fellow find the locus formed GEOMETRY given circle given square GRAMMAR GREEK half Hence HISTORY hypotenuse inscribed internal intersection Introduction JOHN Join late Fellow LATIN Lecturer length LESSONS Let ABC London Maps Master Mathematical maximum meet middle point Notes numerous Illustrations opposite Oxford parallel passing perpendicular position preparation problem produced Prof Professor PROP prove quadrilateral radius ratio rectangle respectively revised right angle School Second segment semicircle sides Similarly solution straight line Students tangent third Translated triangle ABC Trinity College University vertex
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