Elements of geometry, based on Euclid, books i-iii1876 - 119 sider |
Inni boken
Resultat 1-5 av 37
Side 18
... ( Const . ) , and AF is common to the two triangles DAF , EAF ; The two sides DA , AF are equal to the two sides EA , AF , each to each ; And the base DF is equal to the base EF ( Const . ) ; Therefore the angle DAF is equal to the angle ...
... ( Const . ) , and AF is common to the two triangles DAF , EAF ; The two sides DA , AF are equal to the two sides EA , AF , each to each ; And the base DF is equal to the base EF ( Const . ) ; Therefore the angle DAF is equal to the angle ...
Side 19
... ( Const . ) , and CD common to the two triangles ACD , BCD ; The two sides AC , CD are equal to the two sides BC , CD , each to each ; And the angle ACD is equal to the angle BCD ( Const . ) ; Therefore the base AD is equal to the base DB ...
... ( Const . ) , and CD common to the two triangles ACD , BCD ; The two sides AC , CD are equal to the two sides BC , CD , each to each ; And the angle ACD is equal to the angle BCD ( Const . ) ; Therefore the base AD is equal to the base DB ...
Side 20
... ( 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 FHC , GHC . The two sides FH , HC are equal to the 20 GEOMETRY .
... ( 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 FHC , GHC . The two sides FH , HC are equal to the 20 GEOMETRY .
Side 24
... ( Const . ) , AE , EB are equal to CE , EF , each to each ; And the angle AEB is equal to the angle CEF , because they are opposite vertical angles ( I. 15 ) . Therefore the base AB is equal to the base CF ( I. 4 ) ; And the triangle AEB ...
... ( Const . ) , AE , EB are equal to CE , EF , each to each ; And the angle AEB is equal to the angle CEF , because they are opposite vertical angles ( I. 15 ) . Therefore the base AB is equal to the base CF ( I. 4 ) ; And the triangle AEB ...
Side 28
... ( Const . ) ; Therefore FK is equal to A ( Ax . 1 ) . Again , because the point G is the centre of the circle IILK , GH is equal to GK ( Def . 15 ) . But GH is equal to C ( Const . ) ; Therefore GK is equal to C ( Ax . 1 ) , And FG is ...
... ( Const . ) ; Therefore FK is equal to A ( Ax . 1 ) . Again , because the point G is the centre of the circle IILK , GH is equal to GK ( Def . 15 ) . But GH is equal to C ( Const . ) ; Therefore GK is equal to C ( Ax . 1 ) , And FG is ...
Andre utgaver - Vis alle
Elements of Geometry, Based on Euclid, Bøker 1-3 Edward Atkins Ingen forhåndsvisning tilgjengelig - 2016 |
Vanlige uttrykk og setninger
AB is equal AC and CD adjacent angles alternate angles angle ABC angle ACB angle BAC angle BCD angle contained angle DEF angle EDF angle equal angles BGH angles CBA base BC BC is equal bisect centre circle ABC circumference diagonal diameter double draw equal circles equal to AC equal to twice EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line gnomon greater interior and opposite isosceles triangle less Let ABC Let the straight opposite angles parallel to BC parallelogram perpendicular produced PROOF PROOF.-Because Q.E.D. Proposition rectangle AD rectangle AE rectangle contained remaining angle right angle Const right angles Ax segment semicircle side BC square described square on AC touches the circle triangle ABC triangle DEF twice the rectangle
Populære avsnitt
Side 37 - IF a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles.
Side 13 - 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 be equal. Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC be produced to D and E.
Side 7 - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
Side 17 - If two triangles have two sides of the one equal to two sides of the...
Side 53 - 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 9 - If a straight line meet 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 71 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part.
Side 9 - 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 34 - 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 the interior and opposite upon the same side of the line ; or make the interior angles upon the same side together equal to two right angles ; the two straight lines shall be parallel to one another.
Side 69 - If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line, which is made up of the half and the part produced.