Angle Bisector Theorem

Theorem: Bisector of an angle of a trinagle divides the opposite sides in the ratio of the sides containing the angle.

Prerequisites:
Basic Proportionality Theorem (proof)
Alternate angles property (proof)
Corresponding angles property (proof)
Converse of Isosceles Triangle Theorem (proof)

Proof:

Let ABC be a triangle with AD being the bisector of $\angle A$, meeting BC at D. We need to show that:

$\qquad\quad\dfrac{BD}{DC} = \dfrac{AB}{AC}$

For this, let us draw CE $\parallel$ DA to meet the extended BA at C. Since CE $\parallel$ DA,

$\therefore\quad\;\;\;\angle CAD = \angle ACE\qquad\qquad\quad\text{(Alternate angles)}\qquad\qquad\qquad\qquad\qquad\;\;\:\cdots\text{$(1)$}\\[6pt]
\text{Also, }\;\:\!\angle BAD = \angle AEC\qquad\quad\qquad\text{(Corresponding angles)}\qquad\qquad\qquad\qquad\;\;\cdots\text{$(2)$}\\[6pt]
\text{But, }\;\;\angle BAD = \angle CAD\quad\qquad\qquad\:\!\:\!\!\text{(As AD bisects $\angle A$)}\\[6pt]
\therefore\quad\;\;\;\angle ACE = \angle AEC\quad\qquad\!\:\qquad\text{(from $(1)$ and $(2)$)}\\[6pt]
\Rightarrow\quad\;\; AC = AE\qquad\qquad\qquad\qquad\text{(Converse of Isosceles Triangle Theorem)}\quad\cdots\text{$(3)$}$

Now, in $\triangle BCE$,  DA $\parallel$ CE.  Thus, by Basic Proportionality Theorem,

$\qquad\quad\dfrac{BD}{DC} = \dfrac{BA}{AE}\\[8pt]
\Rightarrow\quad\;\;\dfrac{BD}{DC} = \dfrac{AB}{AC}\qquad\qquad\qquad\;\;\text{(from $(3)$)}$

Q.E.D.

Recommended:
Basic Proportionality Theorem
Midpoint Theorem
Location of centroid