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.

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.

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.

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