Pure mathematics, Volum 11874 |
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Side 94
... describe an equilateral triangle on a given finite straight line . Let AB be the given straight line . It is required to describe an equilateral triangle on AB . E CONSTRUCTION . - From the centre A , at the distance AB , describe the ...
... describe an equilateral triangle on a given finite straight line . Let AB be the given straight line . It is required to describe an equilateral triangle on AB . E CONSTRUCTION . - From the centre A , at the distance AB , describe the ...
Side 95
... describe the equilateral triangle DAB ( Book I. , A DAB e- Prop . 1 ) . K 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 ) . K 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 96
... 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 . AB ...
... 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 . AB ...
Side 103
... 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 104
... 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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Pure Mathematics: Including the Higher Parts of Algebra and Plane ..., Volum 1 Edward Atkins Uten tilgangsbegrensning - 1877 |
Vanlige uttrykk og setninger
a²b a²b² a²x² a³b ab² ab³ ABCD adjacent angles algebraical algebraical quantity angle ABC angle BAC angle BCD angle EDF angle equal base BC BC is equal bisect brackets cent centim centre circle ABC coefficient common Const cube root decimal figures denominator distance divided divisor equation expression exterior angle factor Find the value fraction given rectilineal given straight line greater Hence join kilom Let ABC logarithm metres millig Multiply opposite angles parallel parallelogram perpendicular PROOF.-Because Q. E. D. Proposition quotient ratio rectangle contained remainder right angles segment side BC square on AC square root subtraction term triangle ABC x²y x²y² x³y xy² xy³
Populære avsnitt
Side 116 - Wherefore, if a straight line, &c. QED PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines, make the exterior angle equal to...
Side 103 - 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 113 - 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. either the sides adjacent to the equal...
Side 88 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.
Side 273 - The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles.
Side 94 - 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 271 - If a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle ; the angles which this line makes with the line touching the circle, shall be equal to the angles which are in the alternate segments of the circle.
Side 91 - 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 112 - IMS the greater base shall be greater t/uin the angle contained by the sides equal to them of the other. Let ABC, DEF, be two triangles, which have The two sides AB, AC...
Side 128 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.