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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.

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