Proof:

Given: ABC is a triangle, the bisector of the exterior angle at B and the bisector of angle C intersect each other at D.

To Prove: Angle D is equal to half angle A.

Proof:

Step 1: Draw a rough sketch of triangle ABC.

Step 2: Label the given points and angles on the sketch.
- Let ∠A be the angle at vertex A.
- Let ∠B be the angle at vertex B.
- Let ∠C be the angle at vertex C.
- Let ∠D be the angle at vertex D.

Step 3: Use the Angle Bisector Theorem to establish a relationship between the angles.

Step 3a: The Angle Bisector Theorem states that the angle bisector of an angle in a triangle divides the opposite side into two segments that are proportional to the lengths of the other two sides.

Step 3b: Since D is the intersection point of the bisector of the exterior angle at B and the bisector of angle C, we can conclude that D is the angle bisector of both the exterior angle at B and angle C.

Step 3c: Therefore, we can apply the Angle Bisector Theorem to triangle ABC with respect to angle A and angle B.

Step 3d: According to the Angle Bisector Theorem, we have:
AB/BD = AC/CD

Step 3e: Rearrange the equation to isolate AB:
AB = (BD * AC) / CD

Step 4: Use the Converse of the Angle Bisector Theorem to establish an equality between angles A and D.

Step 4a: The Converse of the Angle Bisector Theorem states that if a line divides two sides of a triangle proportionally, then it is an angle bisector.

Step 4b: Since we have established that D is the angle bisector of both the exterior angle at B and angle C, we can conclude that D is also the angle bisector of angle A.

Step 4c: Therefore, angle D is equal to half of angle A.

Step 5: Thus, we have proved that angle D is equal to half of angle A in triangle ABC.