# ANGLE BISECTOR THEOREM PROOF

## Theorem

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,

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,

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. Apart from the stuff given above, if you need any other  stuff in math, please use our google custom search here.

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