The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
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Αποτελέσματα 1 - 5 από τα 32.
Σελίδα 10
A B 1 G D E و In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because ÅF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal ...
A B 1 G D E و In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because ÅF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal ...
Σελίδα 13
Take any point D in AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC .
Take any point D in AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC .
Σελίδα 14
Take any point D in Ac , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point C is 14 EUCLID'S ELEMENTS .
Take any point D in Ac , and ( I. 3. ) make CE equal to CD , and upon DE describe ( I. 1. ) the equilateral triangle DFE , and join FC ; the straight line FC drawn from the given point C is 14 EUCLID'S ELEMENTS .
Σελίδα 16
It is required to draw a straight line perpendicular to AB from the point C. Take any point D upon the other side of AB , and from the centre C , at the distance CD , describe ( Post . 3. ) the circle EGF meeting , AB in F , G ; and ...
It is required to draw a straight line perpendicular to AB from the point C. Take any point D upon the other side of AB , and from the centre C , at the distance CD , describe ( Post . 3. ) the circle EGF meeting , AB in F , G ; and ...
Σελίδα 18
Take away the common angle ABC , and ( Ax . 3. ) 3 . The remaining angle ABE is equal to the remaining angle ABD , the less to the greater , which is impossible ; therefore 4 . BE is not in the same straight line with BC .
Take away the common angle ABC , and ( Ax . 3. ) 3 . The remaining angle ABE is equal to the remaining angle ABD , the less to the greater , which is impossible ; therefore 4 . BE is not in the same straight line with BC .
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AB is equal ABC is equal ABCD angle ABC angle ACB angle AGH angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line gnomon greater impossible inscribed join less Let ABC Let the straight likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle right angles segment semicircle shown side BC sides square of AC straight line AC Take touches the circle triangle ABC twice the rectangle wherefore whole