The converse of this is also true. If a point lies on the interior of an angle and is equidistant from the sides of the angle, then a line from the angle’s vertex through the point bisects the angle.
D is a point in the interior of angle ∠BAC. The perpendicular distances |DC| and |DB| are equal. Show that AD is the angle bisector of angle ∠BAC (∠BAD≅ ∠CAD).
The converse theorem is as easy to prove as the original theorem, once again using congruent triangles.
When asked to prove a converse theory to one you have just proved, it is often a good idea to follow the same strategy as in the original proof, simply switching what needs to be proven with what is already given.
So in this case, △ABD and △ACD share a common side (AD), have another pair of sides which is given as equal (|DC| = |DB|), and both are right triangles – so they are congruent right triangles using the hypotenuse-leg postulate.
It follows that the line AD is the bisector as the angles ∠BAD and ∠CAD are congruent as corresponding parts of congruent triangles.
(1) AD=AD //Common side, reflexive property of equality
(2) |DB|=|DC| //Given
(3) m∠DCA=m∠DBA=90° //definition of distance.
(4)△ABD≅△ACD //(2),(3), Hypotenuse-Leg Postulate
(6) ∠BAD≅ ∠CAD //Corresponding sides of congruent triangles (CPCTC)