Question

Prove that if a point in the interior of an angle is equidistant from the sides of the angle then it belongs to the angle bisector of the angle (Point on angle bisector converse theorem).

Answer 1

To prove the Point on Angle Bisector Converse Theorem, we can use the concept of congruent triangles. Let the point inside the angle be denoted as P. If the distances from P to the two sides of the angle are equal, then we can draw perpendiculars from P to the two sides of the angle, creating two right triangles. By proving that these two right triangles are congruent, we can conclude that the angles opposite the congruent sides are equal, which means that P lies on the angle bisector.

Step-by-step explanation:

In the given question, we are asked to prove the Point on Angle Bisector Converse Theorem, which states that if a point in the interior of an angle is equidistant from the sides of the angle, then it belongs to the angle bisector of the angle.

To prove this, we can use the concept of congruent triangles. Let the point inside the angle be denoted as P. If the distances from P to the two sides of the angle are equal, then we can draw perpendiculars from P to the two sides of the angle, creating two right triangles. By proving that these two right triangles are congruent, we can conclude that the angles opposite the congruent sides are equal, which means that P lies on the angle bisector.

Therefore, if a point in the interior of an angle is equidistant from the sides of the angle, then it belongs to the angle bisector of the angle.

Answer 2

Step-by-step explanation:

Given: <AOB and an interior point D.

Prove: a point, D, in the interior of an angle is equidistant from the sides of the angle, then it belongs to the angle bisector of the angle.

Join D to A and B, then;

OA ⊥ AD (right angle property)

OB ⊥ BD (right angle property)

AD = BD (angle bisector theorem)

<AOD = <BOD (congruent angle property)

ΔAOD = ΔBOD (congruence property of triangles)

Therefore, point D lies on the bisector of <AOB.