57. If a straight line cut the sides, or the sides produced, of a triangle, the product of three segments in order is equal to the product of the other three segments.
Let ABC be a triangle, and let a straight line be drawn cutting the side BC at D, the side CA at E, and the side AB produced through B at F. Then BD and DC are
called segments of the side BC, and CE and EA are called segments of the side CA, and also AF and FB are called segments of the side AB.
Through A draw a straight line parallel to BC, meeting DF produced at H.
Then the triangles CED and EAH are equiangular to one another; therefore AH is to CD as AE is to EC (VI. 4). Therefore the rectangle AH, EC is equal to the rectangle CD, AE (VL 16).
Again, the triangles FAH and FBD are equiangular to one another; therefore AH is to BD as FA is to FB (VI. 4). Therefore the rectangle AH, FB is equal to the rectangle BD, FA (VI. 16).
Now suppose the straight lines represented by numbers in the manner explained in the notes to the second Book of the Elements. We have then two results which we can express arithmetically: namely, the prooduct AH.EC is equal to the product CD.AE; and the product AH.FB is equal to the product BD.FA.
Therefore, by the principles of arithmetic, the product AH.EC.BD.FA is equal to the product AH.FB.CD.AE, and therefore, by the principles of arithmetic, the product BD.CE.AF is equal to the product DC.EA. FB.
This is the result intended by the enunciation given above. Each product is made by three segments, one from
