You are here: BP HOME > BPG > Euclid: Elementa > fulltext
Euclid: Elementa

Choose languages

Choose images, etc.

Choose languages
Choose display
  • Enable images
  • Enable footnotes
    • Show all footnotes
    • Minimize footnotes
Search-help
Choose specific texts..
Euclid: Elementa
    Click to Expand/Collapse Option Complete text
Click to Expand/Collapse OptionTitle
Click to Expand/Collapse OptionPreface
Click to Expand/Collapse OptionBook I
Click to Expand/Collapse OptionBook ΙI
Click to Expand/Collapse OptionBook IΙΙ
Click to Expand/Collapse OptionBook IV
Click to Expand/Collapse OptionBook V
Click to Expand/Collapse OptionBook VI
Click to Expand/Collapse OptionBook VII
Click to Expand/Collapse OptionBook VIII
Click to Expand/Collapse OptionBook ΙΧ
Click to Expand/Collapse OptionBook Χ
Click to Expand/Collapse OptionBook ΧI
Click to Expand/Collapse OptionBook ΧIΙ
Click to Expand/Collapse OptionBook ΧIΙΙ
Back to library
PROPOSITION 32. 
THEOR. 28. PROPOS. 32. 
第三十二題 
If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle. 
SI circulum tetigerit aliqua recta linea, a contactu autem producatur quædam recta linea circulum secans: Anguli, quos ad contingentem facit, æquales sunt ijs, qui in alternis circuli segmentis consistunt, angulis. 
直線切圜。從切界、任作直線、割圜為兩分。分內、各任為負圜角。其切線與割線、所作兩角。與兩負圜角。交(p. 一七四)互相等。 
For let a straight line EF touch the circle ABCD at the point B,  and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it;  I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB. 
     
     
For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD, and let AD, DC, CB be joined. 
 
 
Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19]  Therefore BA is a diameter of the circle ABCD;  therefore the angle ADB, being an angle in a semicircle, is right. [III. 31]  Therefore the remaining angles BAD, ABD are equal to one right angle. [I. 32]  But the angle ABF is also right;  therefore the angle ABF is equal to the angles BAD, ABD.  Let the angle ABD be subtracted from each;  therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle.  Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22]  But the angles DBF, DBE are also equal to two right angles;  therefore the angles DBF, DBE are equal to the angles BAD, BCD,  of which the angle BAD was proved equal to the angle DBF;  therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle. 
                         
                         
Therefore etc.  Q. E. D. 
   
   
 
Permanent link
Go to Wiki Documentation
Enhet: Det humanistiske fakultet   Utviklet av: IT-seksjonen ved HF
Login