Elements of geometry, based on Euclid, book i1877 |
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Resultat 1-5 av 11
Side 10
... describe circles . ACAB . BC = AB . AC and BC each AB . AC = BC . ..AB = BC CA. The sign signifies " is greater than ... describe an equilateral triangle on a given finite straight line . Let AB be the given straight line . It is ...
... describe circles . ACAB . BC = AB . AC and BC each AB . AC = BC . ..AB = BC CA. The sign signifies " is greater than ... describe an equilateral triangle on a given finite straight line . Let AB be the given straight line . It is ...
Side 11
... describe the equilateral triangle DAB ( Book I. , A DAB e- Prop . 1 ) . Produce the straight lines DA , DB , to E and F ( Post . 2 ) . From the centre B , at the dis- tance BC , describe the circle CGH , meeting DF in G ( Post . 3 ) ...
... describe the equilateral triangle DAB ( Book I. , A DAB e- Prop . 1 ) . Produce the straight lines DA , DB , to E and F ( Post . 2 ) . From the centre B , at the dis- tance BC , describe the circle CGH , meeting DF in G ( Post . 3 ) ...
Side 12
... describe the circle DEF , cutting AB in E ( Post . 3 ) . Then AE shall be equal to C. PROOF . - Because the point A is the centre of the circle AE = AD . DEF , AE is equal to AD ( Def . 15 ) . AD = C . AE and C each = AD . .. AE = C ...
... describe the circle DEF , cutting AB in E ( Post . 3 ) . Then AE shall be equal to C. PROOF . - Because the point A is the centre of the circle AE = AD . DEF , AE is equal to AD ( Def . 15 ) . AD = C . AE and C each = AD . .. AE = C ...
Side 19
... describe the equilateral triangle ABC ( I. 1 ) . Bisect the angle ACB by the straight line CD ( I. 9 ) . Then AB shall be cut into two equal parts in the point D. PROOF . - Because AC is equal to CB A D B ( Const . ) , and CD common to ...
... describe the equilateral triangle ABC ( I. 1 ) . Bisect the angle ACB by the straight line CD ( I. 9 ) . Then AB shall be cut into two equal parts in the point D. PROOF . - Because AC is equal to CB A D B ( Const . ) , and CD common to ...
Side 20
... describe the circle EGF , meet- ing AB in F and G ( Post . 3 ) . Bisect FG in H ( I. 10 ) . Join CF , CH , CG . Then CH shall be perpendicular to AB . PROOF . - Because FH is equal to HG Const . ) , and HC common to the two triangles ...
... describe the circle EGF , meet- ing AB in F and G ( Post . 3 ) . Bisect FG in H ( I. 10 ) . Join CF , CH , CG . Then CH shall be perpendicular to AB . PROOF . - Because FH is equal to HG Const . ) , and HC common to the two triangles ...
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Vanlige uttrykk og setninger
ABC is equal adjacent angles alternate angles angle ABC angle BAC angle BCD angle contained angle EDF angle EGB angle GHD angles BGH angles CBE angles equal bisect centre cloth Const describe the circle diagonal equal sides equal to BC equal triangles equilateral triangle exterior angle Fcap four right angles GHD Ax given point given rectilineal angle given straight line given triangle gram HENRY EVERS interior and opposite isosceles triangle join less Let ABC LL.D meet opposite angles parallel straight lines parallel to BC parallelogram ABCD perpendicular Post 8vo PROOF PROOF.-Because Q. E. D. Proposition rectilineal figure remaining angle right angles Ax side BC sides are opposite sides equal square described square GB third angle trapezium triangle ABC triangle DEF WILLIAM COLLINS
Populære avsnitt
Side 23 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it.
Side 33 - If two triangles have two angles of the one equal to two angles of the other, each to each ; and one side equal to one side, viz.
Side 43 - Parallelograms upon the same base, and between the same parallels, are equal to one another.
Side 15 - The angles at the base of an Isosceles triangle are equal to one another ; and if the equal sides be produced, the angles upon the other side of the base shall also be equal. Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC...
Side 11 - Things which are double of the same, are equal to one another. 7. Things which are halves of the same, are equal to one another.
Side 37 - If a straight line meets two straight lines, so as to " make the two interior angles on the same side of it taken " together less than two right angles...
Side 41 - ... together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 15 - J which the equal sides are opposite, shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE.
Side 55 - IF the square described upon one of 'the sides of a triangle be equal to the squares described upon the other two sides of it ; the angle contained by these two sides is a right angle.
Side 24 - If, at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.