Euclid for beginners, books i. and ii., with simple exercises by F.B. HarveyLongmans, Green, and Company, 1880 - 119 σελίδες |
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Σελίδα xx
... joining two of its opposite angles . The Square , Oblong , Rhombus , and Rhomboid are each of them Parallelograms , as this definition shows . For the terms Rhombus and Rhomboid that of Parallelogram is often used ; and for Oblong the ...
... joining two of its opposite angles . The Square , Oblong , Rhombus , and Rhomboid are each of them Parallelograms , as this definition shows . For the terms Rhombus and Rhomboid that of Parallelogram is often used ; and for Oblong the ...
Σελίδα 6
... join BG and CF. PROOF . - Because in the triangles FAC and GAB we have the sides FA and AC , and their angle FAC , in the former = the sides GA and AB , and their angle GAB , in the latter , each to each ( hyp . and cons . ) , therefore ...
... join BG and CF. PROOF . - Because in the triangles FAC and GAB we have the sides FA and AC , and their angle FAC , in the former = the sides GA and AB , and their angle GAB , in the latter , each to each ( hyp . and cons . ) , therefore ...
Σελίδα 8
... join CD . PROOF . - Because in the triangles DBC and ACB we have the sides DB and BC and their angle DBC , in the former = the sides AC and CB , and their angle ACB , in the latter , each to each ( hyp . and cons . ) , therefore the ...
... join CD . PROOF . - Because in the triangles DBC and ACB we have the sides DB and BC and their angle DBC , in the former = the sides AC and CB , and their angle ACB , in the latter , each to each ( hyp . and cons . ) , therefore the ...
Σελίδα 10
... Join the ver- tices CD . PROOF . Because in the triangle ACD the side AC = the side AD ( hyp . ) , therefore the angle ACD = the angle ADC ( I. 5 ) . But the angle ACD is greater than the angle BCD ( ax . 9 ) ; there- fore the angle ADC ...
... Join the ver- tices CD . PROOF . Because in the triangle ACD the side AC = the side AD ( hyp . ) , therefore the angle ACD = the angle ADC ( I. 5 ) . But the angle ACD is greater than the angle BCD ( ax . 9 ) ; there- fore the angle ADC ...
Σελίδα 11
Euclides, Frederick Burn Harvey. CONSTRUCTION . - 1 . Join the vertices CD . 2. Produce AC and AD to E and F respectively . PROOF . - Because in the triangle ACD , the side AC - the side AD ( hyp . ) and these sides are produced to E and ...
Euclides, Frederick Burn Harvey. CONSTRUCTION . - 1 . Join the vertices CD . 2. Produce AC and AD to E and F respectively . PROOF . - Because in the triangle ACD , the side AC - the side AD ( hyp . ) and these sides are produced to E and ...
Άλλες εκδόσεις - Προβολή όλων
Euclid for Beginners, Books I. and II., with Simple Exercises by F.B. Harvey Euclides Δεν υπάρχει διαθέσιμη προεπισκόπηση - 2016 |
Συχνά εμφανιζόμενοι όροι και φράσεις
ABC and ABD AC and CD adjacent angles alternate angle angle ABC angle ACB angle AGH angle BAC angle CEB angle DEF angle EDF angle GHD Arithmetic BA and AC base BC Beginners bisected CONSTRUCTION.-1 crown 8vo Dictionary double the square draw Edition English Grammar English History equilateral Euclid exterior angle Gallic War Geography given straight line gnomon greater Greek half a right i.e. the angle interior and opposite join Latin Let ABC line be divided LONGMANS Manual note 2 def opposite angle parallel parallelogram post 8vo produced PROOF.-Because Proposition proved Q. E. D. Exercise Q. E. D. PROP rectangle contained rectilineal figure right angles School side AB side AC small 8vo square on AC Stepping-Stone straight line CD THEOREM triangle ABC twice the rect twice the rectangle vols Wherefore
Δημοφιλή αποσπάσματα
Σελίδα 48 - 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.
Σελίδα 88 - If a straight line be divided into two equal parts, and also into two unequal parts ; the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Σελίδα 14 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Σελίδα 36 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Σελίδα 64 - 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.
Σελίδα 108 - In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B an acute angle; and on BC one of the sides containing it, let fall the perpendicular...
Σελίδα 47 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line.
Σελίδα 104 - 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.
Σελίδα 52 - The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel.
Σελίδα 20 - If, at a point in a straight line, two other straight lines upon 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.