I WILL GIVE BRAINLIEST Prove that any point that belongs to the angle bisector of an angle is equidistant from the sides of the angle (Point on Angle bisector).

I think my answer is what you'd like:Draw the angle and the angle bisector. Take any point on the bisector and drop perpendiculars from that point onto the two arms of the angle. Now you have two triangles formed by the two arms, the bisector and the 2 perpendiculars that you drew. The two triangles are congruent because two angles are equal (the bisected angle) and the right angles and one side is common. Hence the perpendiculars are equal in length and thus the angle bisector is equidistant from the sides of the angle.Proof:Lets have a line divided into two parts by a point S. Lets construct a ray from S that has an angle of alpha with the left ray beginning with S. Lets construct another angle of measure alpha, this time using the right ray as a side of our angle. If alpha is arbitrary, then the angle between them is also arbitrary. Lets construct a line perpendicular to line on which S lies that goes through S. Lets choose an arbitrary point on the line P (the point should be above S). Lets construct two lines going through P that are perpendicular to the sides of angles we have already constructed. Lets name the intersection points X and Y. I claim that when we draw a line segment between X and Y we get a segment perpendicular to our base line (the one with S). I, further claim that the triangle PXY is isoceles. (We use the fact that the left our two identical angles is congruent to its alternate interior angle. We also know that the PXY is equal to 90-alpha degrees We can use the same deduction to show that PYX is 90-alpha degrees.) Therefore the distance between PX and PY is identical.Hope that helps!!!!!