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 .

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,

the angle ACE equals the angle CAD  because alternate angles are equal. Euclid Book 1, Proposition 29.

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.

 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.

Therefore we have proven:

Book 6, Proposition 3
If an angle of a triangle is bisected by a straight line cutting the base, then the segments of the base have the same ratio as the remaining sides of the triangle;

https://www.math.ubc.ca/~cass/euclid/book6/images/bookVI-prop3.html
(this page is incorrectly identified as Book 5 at the top.)  Or see
http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html  More on Euclid

And to repeat the all of the above, for those who like it simply put,