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Euclid: Elementa
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PROPOSITION 35. 
THEOR. 29. PROPOS. 35. 
第三十五題 
If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other. 
SI in circulo duæ rectæ lineæ sese mutuo secuerint, rectangulum comprehensum sub segmentis vnius, æquale est ei, quod sub segmentis alterius comprehenditur, rectangulo. 
圜內兩直線。交而相分。各兩分線矩內直角形、等。 
For in the circle ABCD let the two straight lines AC, BD cut one another at the point E;  I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB. 
   
   
If now AC, BD are through the centre, so that E is the centre of the circle ABCD,  it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB. 
   
   
Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F;  from F let FG, FH be drawn perpendicular to the straight lines AC, DB,  and let FB, FC, FE be joined. 
     
     
Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles,  it also bisects it; [III. 3]  therefore AG is equal to GC.  Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E,  the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5]  Let the square on GF be added;  therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF.  But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47]  therefore the rectangle AE, EC together with the square on FE is equal to the square on FC.  And FC is equal to FB;  therefore the rectangle AE, EC together with the square on EF is equal to the square on FB.  For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB.  But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB;  therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE.  Let the square on FE be subtracted from each;  therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB. 
                               
                               
Therefore etc.   
   
   
 
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