Angle Bisector Theorem

The angle bisector theorem concerns about the relevant lengths of two segments which is divided by a line which bisects the opposite angle. Their relevant lengths are equated to relevant lengths of the other two sides.

Angle Bisector Theorem

Theorem

In the triangle ABC, the angle bisector intersects side BC at the point D. Then, According to Angle bisector theorem, the ratio of the line segment BD to DC equals to the ratio of length of the side AB to AC.

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}\)

Conversely, when a point D on the side BC divides BC in the ratio similar to the sides AC and AB, then the angle bisector of ∠ A is AD. Hence, according to the theorem, if D lies on the side BC, then,

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |Sin\angle DAB}{\left | AC \right |Sin\angle DAC}\)

If D is external to the side BC, directed angles and directed line segments are required to be applied in the calculation.

Angle bisector theorem is applied when side lengths and angle bisectors are known.

Proof

Use law of sines on triangles ABD and ACD in the above figure.

\(\frac{AB}{BD}=\frac{sin\angle BDA}{sin\angle BAD}\)

\(\frac{AC}{DC}=\frac{sin\angle ADC}{sin\angle DAC}\)

The angles ∠ ADC and ∠ BDA make a linear pair which and hence called as adjacent supplementary angles. Since the sines of supplementary angles are equal,

Sin ∠ BDA = Sin ∠ ADC

∠ DAC and ∠ BAD are equal. Hence, the RHS of the equations 1 and 2 are equal, therefore LHS must also be equal.

\(\frac{\left | BD \right |}{\left | DC \right |}=\frac{\left | AB \right |}{\left | AC \right |}\)

If the angles ∠ DAC and ∠ BAD are not equal, the equation 1 and equation 2 can be written as:

\(\frac{\left | AB \right |}{\left | BD \right |}\) sin ∠ BAD = sin∠ BDA

\(\frac{\left | AC \right |}{\left | DC \right |}\) sin ∠ DAC = sin∠ ADC

Angles ∠ ADC and ∠ BDA are supplementary, hence the RHS of the equations are still equal. Hence, we get

\(\frac{\left | AB \right |}{\left | BD \right |}\) sin ∠BAD = \(\frac{\left | AC \right |}{\left | DC \right |}\) sin ∠DAC

This rearranges to generalized view of the theorem.


Practise This Question

If z is a complex number such that z2 = (¯z)2,then