There exist many different ways of proving the angle bisector theorem. A few of them are shown below.
Proof using similar triangles As shown in the accompanying animation, the theorem can be proved using similar triangles. In the version illustrated here, the triangle \triangle ABC gets reflected across a line that is perpendicular to the angle bisector AD, resulting in the triangle \triangle A B_2 C_2 with bisector AD_2. The fact that the bisection-produced angles \angle BAD and \angle CAD are equal means that BA C_2 and CA B_2 are straight lines. This allows the construction of triangle \triangle C_2BC that is similar to \triangle ABD. Because the ratios between corresponding sides of similar triangles are all equal, it follows that |AB|/|AC_2| = |BD|/|CD|. However, AC_2 was constructed as a reflection of the line AC, and so those two lines are of equal length. Therefore, |AB|/|AC| = |BD|/|CD|, yielding the result stated by the theorem.
Proof using law of sines In the above diagram, use the
law of sines on triangles and : {{NumBlk|:|{\frac } = {\frac {\sin \angle ADB} {\sin \angle DAB}} |}} {{NumBlk|:|{\frac } = {\frac {\sin \angle ADC} {\sin \angle DAC}} |}} Angles and form a linear pair, that is, they are adjacent
supplementary angles. Since supplementary angles have equal sines, : = {\sin \angle ADC}. Angles and are equal. Therefore, the right hand sides of equations () and () are equal, so their left hand sides must also be equal. :{\frac }={\frac }, which is the angle bisector theorem. If angles are unequal, equations () and () can be re-written as: : {\frac \sin \angle DAB = \sin \angle ADB}, : {\frac \sin \angle DAC = \sin \angle ADC}. Angles are still supplementary, so the right hand sides of these equations are still equal, so we obtain: : {\frac \sin \angle DAB = \frac \sin \angle DAC}, which rearranges to the "generalized" version of the theorem.
Proof using triangle altitudes Let be a point on the line , not equal to or and such that is not an
altitude of triangle . Let be the base (foot) of the altitude in the triangle through and let be the base of the altitude in the triangle through . Then, if is strictly between and , one and only one of or lies inside and it can be assumed
without loss of generality that does. This case is depicted in the adjacent diagram. If lies outside of segment , then neither nor lies inside the triangle. are right angles, while the angles are congruent if lies on the segment (that is, between and ) and they are identical in the other cases being considered, so the triangles are similar (AAA), which implies that: :{\frac }= {\frac } = \frac = \sin \angle \ BAD \text{ and } \frac = \sin \angle \ DAC, and the generalized form follows.
Proof using isosceles triangles Construct point D' on the bisector such that \triangle ABD\sim\triangle ACD'. We aim to show that |CD|=|CD'|. In the case where D' lies on \overline{AD}, we have that \angle CD'D=180^\circ-\angle CD'A=180^\circ-\angle BDA=\angle CDD',and in the case where D' does not lie on \overline{AD}, we have that \angle CD'D=\angle CD'A=\angle BDA=\angle CDD'.Either way, \triangle CDD' is isosceles, implying that |CD|=|CD'|. Therefore, \frac=\frac=\frac,which was the desired result.
Proof using triangle areas A quick proof can be obtained by looking at the ratio of the areas of the two triangles , which are created by the angle bisector in . Computing those areas twice using
different formulas, that is \tfrac{1}{2}gh with base g and altitude and \tfrac{1}{2}ab\sin(\gamma) with sides and their enclosed angle , will yield the desired result. Let denote the height of the triangles on base and \alpha be half of the angle in . Then : \frac = \frac{\frac{1}{2}|BD|h}{\frac{1}{2}|CD|h} = \frac and : \frac = \frac{\frac{1}{2}|AB||AD|\sin(\alpha)}{\frac{1}{2}|AC||AD|\sin(\alpha)} = \frac yields : \frac = \frac. == Length of the angle bisector ==