Geometrical Exercises for BeginnersMacmillan, 1882 - 203 sider |
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Side 2
... 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 . Note . The converse of this very important theorem may be ...
... 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 . Note . The converse of this very important theorem may be ...
Side 2
... 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 . Note . The converse of this very important theorem may be ...
... 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 . Note . The converse of this very important theorem may be ...
Side 6
... Similarly the triangles BOL , NER , and BOC , ERF , are equal ; therefore the triangle DEF is equal to the figure ABCOA . But the triangles DEF , ABC are equal ( I. 38 ) ; therefore the triangle ABC is equal to the figure ABCO , which ...
... Similarly the triangles BOL , NER , and BOC , ERF , are equal ; therefore the triangle DEF is equal to the figure ABCOA . But the triangles DEF , ABC are equal ( I. 38 ) ; therefore the triangle ABC is equal to the figure ABCO , which ...
Side 8
... Similarly the angles ACD , ABC are equal . PROP . 7. The area of a triangle is equal to half the rectangle contained by its base and altitude . E A F B D Let ABC be any triangle ; and from the vertex , A , draw AD perpendicular to the ...
... Similarly the angles ACD , ABC are equal . PROP . 7. The area of a triangle is equal to half the rectangle contained by its base and altitude . E A F B D Let ABC be any triangle ; and from the vertex , A , draw AD perpendicular to the ...
Side 14
... Similarly Since the angle ADB is right , AB2 = AD2 + BD2 ( I. 47 ) . ACAD + CD3 ; .. by subtraction ABa — AC2 = BD2 — CD3 . PROP . 12. The sum of the squares of the sides of a triangle is equal to double the square of the bisector of ...
... Similarly Since the angle ADB is right , AB2 = AD2 + BD2 ( I. 47 ) . ACAD + CD3 ; .. by subtraction ABa — AC2 = BD2 — CD3 . PROP . 12. The sum of the squares of the sides of a triangle is equal to double the square of the bisector of ...
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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
AD² angle BAC angle BCD angle equal Assistant-Master base angles base BC BD² Bisect Cambridge centre chord circumference circumscribing circle Clifton College Conic Sections Consider Prop construct Crown 8vo draw ELEMENTARY TREATISE English equal to half escribed circle Eton College Extra fcap Fellow of St find the locus fixed point GEOMETRY given base given circle given difference given ratio given square given straight line Given the base given vertical angle GRAMMAR GREEK half a right half the given Hence HISTORY hypotenuse inscribed internal bisector intersection isosceles J. P. MAHAFFY John's College late Fellow LATIN Let ABC Mathematical meet middle point nine-point circle numerous Illustrations Owens College Oxford parallel perpendicular perpendicular to BC Professor prove quadrilateral R. C. JEBB radius rectangle rectangle contained right angle right-angled triangle School segment semicircle tangent triangle ABC Trinity College vertex W. K. CLIFFORD
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