The internal bisector of an angle …

The internal (external) bisector of an angle of a triangle divides the opposite side internally (externally) in the ratio of the corresponding sides containing the angle.

Case (i) (Internally) : 

Given : In ΔABC, AD is the internal bisector of ∠BAC which meets BC at D. 

To prove : BD/DC  =  AB/AC

Construction : Draw CE ∥ DA to meet BA produced at E.

Proof (Internally) : 

Because CE ∥ DA and AC is the transversal, we have 

∠DAC  =  ∠ACE (alternate angles) -----(1)

and

∠BAD  =  ∠AEC (corresponding angles) -----(2)

Because AD is the angle bisector of ∠A, 

∠BAD  =  ∠DAC -----(3)

From (1), (2) and (3), we have 

∠ACE  =  ∠AEC

Thus in ΔACE, we have 

AE  =  AC

(Sides opposite to equal angles are equal)

Now, in ΔBCE we have, CE ∥ DA. 

By Thales Theorem, 

BD/DC  =  BA/AE

Because AE  =  AC, 

BD/DC  =  AB/AC

Hence the theorem.

Case (i) (Externally) : 

Given : In ΔABC, AD is the external bisector of ∠BAC and intersects BC produced at D. 

To prove : BD/DC  =  AB/AC

Construction : Draw CE ∥ DA meeting AB at E.

Proof (Externally) :

Because CE ∥ DA and AC is a transversal, we have 

∠ECA  =  ∠CAD (alternate angles) -----(1)

Also, CE ∥ DA and BP is a transversal, we have 

∠CEA  =  ∠DAP (corresponding angles) -----(2)

But AD is the bisector of ∠CAP, 

∠CAD  =  ∠DAP -----(3)

From (1), (2) and (3), we have 

∠CEA  =  ∠ECA

(Sides opposite to equal angles are equal)

In ΔBDA, we have EC ∥ AD. 

By Thales Theorem, 

BD/DC  =  BA/AE

Because AE  =  AC, 

BD/DC  =  BA/AC

Hence the theorem.