Elements of geometry, based on Euclid, books i-iii1876 - 119 sider |
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Side 16
... equal to one another ; And their sides CB , DB , terminated in the extremity B of the base , likewise equal to one another . CASE I. - Let the vertex of each triangle Bbe without the other triangle . CONSTRUCTION . — Join CD . PROOF ...
... equal to one another ; And their sides CB , DB , terminated in the extremity B of the base , likewise equal to one another . CASE I. - Let the vertex of each triangle Bbe without the other triangle . CONSTRUCTION . — Join CD . PROOF ...
Side 19
... CD ( I. 9 ) . Then AB shall be cut into two equal parts in the point D. D B PROOF . Because AC is equal to CB A ( Const . ) , and CD common to the two triangles ACD , BCD ; The two sides AC , CD are equal to the two sides BC , CD , each ...
... CD ( I. 9 ) . Then AB shall be cut into two equal parts in the point D. D B PROOF . Because AC is equal to CB A ( Const . ) , and CD common to the two triangles ACD , BCD ; The two sides AC , CD are equal to the two sides BC , CD , each ...
Side 20
... equal to the angle EBA ( Def . 10 ) . Therefore the angle DBE is equal to the angle CBE . The CBE . less to the greater ; which is impossible . CD ... CD , describe the circle EGF , meet- ing AB in F and G ( Post . 3 ) . Bisect FG in H ( I.
... equal to the angle EBA ( Def . 10 ) . Therefore the angle DBE is equal to the angle CBE . The CBE . less to the greater ; which is impossible . CD ... CD , describe the circle EGF , meet- ing AB in F and G ( Post . 3 ) . Bisect FG in H ( I.
Side 21
... equal to the base CG ( Def . 15 ) ; Therefore the angle CHF is equal to the angle CHG ( I. 8 ) , .. adjacent and ... CD , upon one side of it , the angles CBA , ABD . These angles shall either be two right angles , or shall to- gether be ...
... equal to the base CG ( Def . 15 ) ; Therefore the angle CHF is equal to the angle CHG ( I. 8 ) , .. adjacent and ... CD , upon one side of it , the angles CBA , ABD . These angles shall either be two right angles , or shall to- gether be ...
Side 23
... equal . Let the two straight lines AB , CD cut one another in the point E. The angle AEC shall be equal to C angle DEB , and the angle CEB to the angle AED . PROOF . - Because the straight line AE makes with CD , the angles CEA , A E B ...
... equal . Let the two straight lines AB , CD cut one another in the point E. The angle AEC shall be equal to C angle DEB , and the angle CEB to the angle AED . PROOF . - Because the straight line AE makes with CD , the angles CEA , A E B ...
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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.