Geometrical Problems Deducible from the First Six Books of Euclid, Arranged and Solved: To which is Added an Appendix Containing the Elements of Plane Trigonometry ...J. Smith, 1821 - 438 sider |
Fra bogen
Resultater 1-5 af 100
Side 4
... the same man- ner , if pC be joined , it may be shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap ...
... the same man- ner , if pC be joined , it may be shewn that Ap = p C. Hence AP and BP together are equal to BC , and Ap , pB are equal to Cp , pB . Now ( Eucl . i . 20. ) BC is less than Bp , PC , and therefore AP , PB are less than Ap ...
Side 11
... BC ; and therefore parallel to each other ; and draw DK parallel to AB . B Then because GD is parallel to HE one of the sides of the triangle AHE , AG : GH :: AD : DE ; hence AG = GH . For the same reason DL = LM . But DM being parallel ...
... BC ; and therefore parallel to each other ; and draw DK parallel to AB . B Then because GD is parallel to HE one of the sides of the triangle AHE , AG : GH :: AD : DE ; hence AG = GH . For the same reason DL = LM . But DM being parallel ...
Side 12
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC BG :: FD : FG . But FE being parallel to BC , ( Eucl . vi . 2. ) AB BC : AF ...
... BC is parallel to FD , the angle BCG is equal to GDF and the vertically opposite angles at G are equal ; therefore the triangles DGF , BGC are similar , and BC BG :: FD : FG . But FE being parallel to BC , ( Eucl . vi . 2. ) AB BC : AF ...
Side 21
... equal to a given line . Let AB , AC , BC be the three lines given in position , take AD = the given line , and making with AB an angle equal to one of the given angles . Through D draw Dba parallel to AB , and meeting AC and BC in a and ...
... equal to a given line . Let AB , AC , BC be the three lines given in position , take AD = the given line , and making with AB an angle equal to one of the given angles . Through D draw Dba parallel to AB , and meeting AC and BC in a and ...
Side 36
... BC CD in the given ratio ; and from the points E and B D , in which CD cuts the semicircle , draw EB , AD to the extremities of the diameter . CD is the line required . Since the angles EDA , EBA in the same segment are equal , and the ...
... BC CD in the given ratio ; and from the points E and B D , in which CD cuts the semicircle , draw EB , AD to the extremities of the diameter . CD is the line required . Since the angles EDA , EBA in the same segment are equal , and the ...
Indhold
1 | |
157 | |
158 | |
164 | |
170 | |
171 | |
177 | |
179 | |
262 | |
264 | |
267 | |
273 | |
277 | |
283 | |
289 | |
295 | |
185 | |
191 | |
193 | |
199 | |
203 | |
209 | |
212 | |
218 | |
220 | |
225 | |
231 | |
232 | |
237 | |
243 | |
249 | |
255 | |
256 | |
301 | |
303 | |
311 | |
319 | |
18 | |
22 | |
49 | |
72 | |
108 | |
142 | |
187 | |
209 | |
227 | |
268 | |
293 | |
399 | |
Andre udgaver - Se alle
Almindelige termer og sætninger
ABCD angle ABC angle equal BC is equal centre chord circle cutting circumference construct the triangle cosecant cosine describe a circle divided equation equiangular Eucl extremities find the value given angle given circle given difference given in position given line given point given ratio given rectangle given square given straight line Hence inscribed intercepted isosceles triangle Join AE least common multiple Let AB Let ABC let fall line given line joining lines be drawn lines drawn mean proportional meeting opposite side parallel to AC parallelogram pendicular point of intersection produced quadrant radius rectangle contained right angles right-angled triangle segments semicircle shewn sine squares of AC tang tangent transposition trapezium triangle ABC triangle required vertex vertical angle whence
Populære passager
Side i - IF a straight line be divided into two equal, and also into two unequal parts ; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.
Side xi - IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle,. shall be equal to the square of the line which touches it.
Side 319 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Side 150 - Iff a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other...
Side 204 - FC are equal to one another : wherefore the circle described from the centre F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC.
Side 115 - If from a point, without a parallelogram, there be drawn two straight lines to the extremities of the two opposite sides, between which, when produced, the point does not lie, the difference of the triangles thus formed is equal to half the parallelogram. Ex. 2. The two triangles, formed by drawing straight lines from any point within a parallelogram to the extremities of its opposite sides, are together half of the parallelogram.
Side 14 - In one of the given equations obtain the value of one of the unknown quantities in terms of the other unknown quantity; Substitute this value in the other equation and solve.
Side 291 - AB describe a segment of a circle containing an angle equal to the given angle, (in.
Side 297 - Given the vertical angle, the difference of the two sides containing it, and the difference of the segments of the base made by a perpendicular from the vertex ; construct the triangle.
Side 90 - If from any point in the base of an isosceles triangle perpendiculars...