The Angle Bisector Theorem And Its Converse: Proving Angle Bisectors In Triangles

Converse of the Angle Bisector Theorem

If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angleIf AP = BP, then CP bisects ACB

The Angle Bisector Theorem states that if a ray bisects an angle of a triangle, it divides the opposite side into two segments that are proportional to the other two sides of the triangle.

The converse of the Angle Bisector Theorem is also true. That is, if a line segment splits a triangle’s opposite side into two portions that are proportional to the triangle’s other two sides, then that line segment is the bisector of the corresponding angle opposite to the divided side.

More formally, let ∆ABC be a triangle with D lying on BC such that BD/DC=AB/AC. Then the line segment AD is the bisector of the angle at vertex A.

This converse theorem is useful in problems where we are given the ratio of two sides of a triangle and need to prove that a certain line segment is an angle bisector. We can use the ratio of the sides to show that the line segment divides the opposite side in the same ratio, which in turn proves that it is an angle bisector.

For example, if we are given that BD/DC=AB/AC in ∆ABC, and we need to prove that the angle bisector of A also passes through the incenter of the triangle, we can use the converse of the Angle Bisector Theorem to show that the incenter lies on the angle bisector by showing that AD is an angle bisector.

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