Question

FAN & MEDAL

Answer 1

ok wat do you need help with

Answer 2

How do you prove that a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects?

Answer 3

Proof:
Let [AB] be a given segment. Let M be an arbitrary point in space such that AM = BM.
Project M onto (d) and call H the intersection point of (d) and (AB).
Consider the triangles AMH and BMH.
These two triangles have the angle HBM = the angle HAM (Because the triangle ABM is isosceles in M), and the angle BHM = the angle AHM (= 90 degrees). Therefore, the angle AMH = the angle BMH (two of the angles are the same in both triangles, so the other must be the same)
These triangles also have AM = BM, they also have a common side which is [MH].
In summary, we showed that the triangles have two equal sides, and the angles between those sides are also equal. Therefore these triangles are congruent. Therefore AH = BH and thus H is the midpoint of [AB]. Hence, (d) is on the perpendicular bisector of the [AB].
Hope this helped!!! : )

Answer 4

ill come back 2 you @firelord

Answer 5

Sorry took a while to type this up!!!

Answer 6

Thanks!