Angle Bisector Theorem
What is Angle Bisector Theorem?
Angle Bisector Theorem states that "In a triangle, the angle bisector of any angle will divide the opposite side in the ratio of the sides containing the angle".
Consider the figure below, in which \(AD\) is the bisector of \(\angle A\).
Angle Bisector Theorem tells us that \(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)
Proof of Angle Bisector Theorem:
Consider \(\Delta ABC\),
Given: \(AD\) is the bisector of \(\angle A\)
To Prove: \(\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}\)
Construction: Draw a ray \(CX\) parallel to \(AD\), and extend \(BA\) to intersect this ray in \(E\)
Proof:
By the Basic Proportionality Theorem, We know that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
\(\therefore \) In \(\Delta CBE\), \(DA\parallel CE\)
\[{\text{}} \Rightarrow \frac{{BD}}{{DC}} = \frac{{BA}}{{AE}}....(1)\]
Now, we only need to prove that \(AE = AC\).
For that, consider the figure above once again, with the angles marked, as shown below:
Since \(DA\parallel CE\), We have:

\(\angle1\) = \(\angle 2\) (corresponding angles)

\(\angle 3\) = \(\angle 4\) (alternate interior angles)
But \(AD\) is the bisector of \(\angle BAC\), which means that \(\angle 1\) = \(\angle 3\).
Thus, \(\angle 2\) = \(\angle 4\).
\[ \Rightarrow AC = AE{\text{ }}({\text{Sides opposite to equal angles are equal}})....{\text{(2)}}\]
Finally, from (1) and (2),
\[\boxed{\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}}\]
The angle bisector \(AD\) thus divides the opposite side \(BC\) in the ratio of the sides \((AB\) and \(AC)\) containing that angle.
Hence Proved.
✍Note: Angle bisector theorem is an important consequence of BPT.
Challenge: In \(\Delta PQR\), \(E\) is a point on \(QR\) such that \(PE\) is the bisector of \(\angle P\)
\(PQ = 4\) units, \(QE = 2\) units, \(ER = 3\) units.
Find \(PR\).
⚡Tip: By Angle Bisector Theorem, \(\frac{{QE}}{{ER}} = \frac{{PQ}}{{PR}}\)