In Fig. 6.63, D is a point on side BC of Δ ABC such that.

  Construct a line CM which meets BA extended up to AM so that AM = AC

This means, ∠AMC = ∠ACM     (Angles opposite to equal sides)

     (Given)

or,

BECAUSE AM = AC

Hence,

ΔABD ΔMBC

Hence, AD parallel MC

This means,

∠DAC = ∠ACM     (Alternate angles)

Since,

∠AMC = ∠ACM

Hence,

∠DAC = ∠BDA

Hence, AD is the bisector of ∠BAC