Elements of Geometry, Briefly, Yet Plainly Demonstrated by Edmund StoneD. Midwinter, 1728 |
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Seite 22
... continue ། Now because CE EA , and EF8EB , h15.1 . and Angle FEC BEA ; the Angle ECF fhall be EAB . In like manner , the Angle ICH therefore the whole Angle ACD BCG ) is greater than either CAB Q. E. D. ABH k 15.1 . ( that is , 19 ax ...
... continue ། Now because CE EA , and EF8EB , h15.1 . and Angle FEC BEA ; the Angle ECF fhall be EAB . In like manner , the Angle ICH therefore the whole Angle ACD BCG ) is greater than either CAB Q. E. D. ABH k 15.1 . ( that is , 19 ax ...
Seite 24
... Continue out the Line BA , and take AD = AC , and draw the Line DC ; then fhall the Angle D bed equal to ACD ; whence the whole Angle BCD D : and fo BD f ( 8 BA + AC ) BC . Q.E. D. PROP . A D E B • BC . XXI . If from the Extreme Points ...
... Continue out the Line BA , and take AD = AC , and draw the Line DC ; then fhall the Angle D bed equal to ACD ; whence the whole Angle BCD D : and fo BD f ( 8 BA + AC ) BC . Q.E. D. PROP . A D E B • BC . XXI . If from the Extreme Points ...
Seite 40
... Continue out DG to meet a right Line drawn from I in the Point K. Thro ' K draw KL pa- rallel to GH , which let EF , IH continued out meet in M , L. Then fhall FL be the Pgr . fought . e For the Pgr . FL- FD = B , and the Ang . MFH GFE ...
... Continue out DG to meet a right Line drawn from I in the Point K. Thro ' K draw KL pa- rallel to GH , which let EF , IH continued out meet in M , L. Then fhall FL be the Pgr . fought . e For the Pgr . FL- FD = B , and the Ang . MFH GFE ...
Seite 53
... continue out the Line AB fo that BD = CB , and describe AF the Square of AD , and conftruct the double Square . " a = ' b = Since CB BD = GK KN , and PR = RO , the Rectang . CK fhall be - Rectang . BN , and the Rectang . GR = Rectang ...
... continue out the Line AB fo that BD = CB , and describe AF the Square of AD , and conftruct the double Square . " a = ' b = Since CB BD = GK KN , and PR = RO , the Rectang . CK fhall be - Rectang . BN , and the Rectang . GR = Rectang ...
Seite 56
... continue out HG to I , then a the Rectang . DHEA 2 2 EF EB2 e 2 - ΕΒ е BA + EA . Therefore DH = BA Square AC . Take away the common Rectang . AI , and there will remain the Square AH GC ; that is , AG2 = AB × BG . Q. E. D. 2 f SCHOLIU M ...
... continue out HG to I , then a the Rectang . DHEA 2 2 EF EB2 e 2 - ΕΒ е BA + EA . Therefore DH = BA Square AC . Take away the common Rectang . AI , and there will remain the Square AH GC ; that is , AG2 = AB × BG . Q. E. D. 2 f SCHOLIU M ...
Häufige Begriffe und Wortgruppen
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Beliebte Passagen
Seite 31 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Seite 145 - Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional : And triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.
Seite 33 - ABD, is equal* to two right angles, «13. 1. therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles.
Seite 27 - If two right-angled triangles have their hypotenuses equal, and one side of the one equal to one side of the other, the triangles are congruent.
Seite 31 - The three angles of any triangle taken together are equal to the three angles of any other triangle taken together. From whence it follows, 2.
Seite 11 - That a straight line may be drawn from any point to any other point. 2. That a straight line may be produced to any length in a straight line.