prove that the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle

Given: In ∆ABC, AD is the angle bisector

Construction: Draw a line from C parallel to AD to intersect BA produced of E

Now ∠1 = ∠2        (AD is angle bisector)  ..... (1)

∠2 = ∠3       (Alternate opposite angles as AD ║ EC and AC is the transversal)  ........ (2)

∠1 = ∠4          ( corresponding angles as AD ║ EC and BE is the transversal)  ..... (3)

from (1), (2), (3)

∠3 = ∠4

Thus in ∆ACE

AE = AC (sides opposite to equal angles)  ....... (4)

Now in ∆ABD and ∆EBC

∠ABD = ∠EBC           (Common)

∠BAD = ∠BEC           (from (3))

⇒ ∆ABD ~ ∆EBC