Euclid for beginners, books i. and ii., with simple exercises by F.B. Harvey |
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Page 81
... gnomon AKG . 2. Similarly the parallelogram HG , about the diameter , to- gether with the complements AF and FC , forms the gnomon EHC . 3. The gnomon is briefly expressed by the letters at the oppo- site angles of the parallelograms ...
... gnomon AKG . 2. Similarly the parallelogram HG , about the diameter , to- gether with the complements AF and FC , forms the gnomon EHC . 3. The gnomon is briefly expressed by the letters at the oppo- site angles of the parallelograms ...
Page 90
... DM ( ax . 2 ) , the whole DF . AL ( I. 36 ) , therefore AL DF ( ax . 1 ) , and therefore AL and CH = DF and CH ( ax . 2. ) , i.e. the rect . AH = the gnomon CMG . But because AH is the rect . AD , DH 90 EUCLID BOOK II .
... DM ( ax . 2 ) , the whole DF . AL ( I. 36 ) , therefore AL DF ( ax . 1 ) , and therefore AL and CH = DF and CH ( ax . 2. ) , i.e. the rect . AH = the gnomon CMG . But because AH is the rect . AD , DH 90 EUCLID BOOK II .
Page 91
... gnomon CMG = the rect . AD , DB ( ax . 1 ) ; therefore also the gnomon CMG together with LG , i.e. the whole figure CEFB = { the rect . AD , DB , together with LG ; But because LG is the square on LH ( II.4 , Cor . ) and LH = CD ( I. 34 ) ...
... gnomon CMG = the rect . AD , DB ( ax . 1 ) ; therefore also the gnomon CMG together with LG , i.e. the whole figure CEFB = { the rect . AD , DB , together with LG ; But because LG is the square on LH ( II.4 , Cor . ) and LH = CD ( I. 34 ) ...
Page 92
... draw AK parallel to CL or DM . PROOF.- Because ALCH ( I. 36 ) , and CH HF ( I. 43 ) , therefore AL HF ( ax . 1 ) ; and therefore AL and CM = HF and CM , i.e. AM the gnomon CMG . But because AM is the rect . AD , DM 92 EUCLID , BOOK II .
... draw AK parallel to CL or DM . PROOF.- Because ALCH ( I. 36 ) , and CH HF ( I. 43 ) , therefore AL HF ( ax . 1 ) ; and therefore AL and CM = HF and CM , i.e. AM the gnomon CMG . But because AM is the rect . AD , DM 92 EUCLID , BOOK II .
Page 94
... gnomon AKF and CK , therefore the gnomon AKF and CK = twice AK ( ax . 1 ) . Next , because AK is the ( II . 4 94 EUCLID , BOOK II .
... gnomon AKF and CK , therefore the gnomon AKF and CK = twice AK ( ax . 1 ) . Next , because AK is the ( II . 4 94 EUCLID , BOOK II .
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Common terms and phrases
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
Popular passages
Page 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.
Page 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.
Page 14 - To draw a straight line at right angles to a given straight line, from a given point in the same.
Page 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.
Page 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.
Page 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...
Page 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.
Page 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.
Page 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.
Page 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.