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Euclid’s Book 6, Proposition 3
If an angle of a triangle is bisected by a straight line cutting the
base, then the remaining sides of the triangle have the same ratio of the segments of the base
Let us prove Euclid’s
theorem in Book 6, proposition 3 which states :
BA is to AC as
BD is to DC . 


Begin with triangle ABC.





Draw Blue Line AD
so that it bisects BAC.




Draw CE through C parallel to DA,





and carry BA through to meet it at E.





Then, since the straight line AC falls upon





the
parallels AD and EC,






And the angle BAD equals the angle DAC by hypothesis since
AD bisects the original angle BAC.





And therefore the angle BAD also equals the
angle ACE.





And since the straight line BAE falls upon the
parallels AD and EC, the exterior angle BAD
equals the interior angle AEC. (Bk
I. Prop. 29, or
BB) 




But the angle ACE was also proved equal to the
angle BAD,
therefore the angle ACE also equals
the angle AEC,





When two angles of a triangle are equal in size the sides
opposite those angles are also equal. So, side AE equals
the side AC.
Book 1, Proposition 6
Book 6, Proposition 2 states :
If a straight line is drawn parallel to one of the sides of
a triangle, then it cuts the sides of the triangle
proportionally; (Book
6, Proposition 2, or
Euclid)
And, since





AD is parallel to EC, Therefore





the two
sides of the triangle BEC are divided proportionally
by AD.
Therefore
BA is to AE as
BD is to DC .
Since,





AE equals AC,




BA is to AC as BD is to DC. 













