$A, B$ বিন্দু থেকে সমদূরবর্তী বিন্দু $AB$ এর লম্বসমদ্বিখন্ডকের ওপর অবস্থিত

It turns out that points on the perpendicular bisector of a segment have a special, important property. Let's look at an arbitrary point $X$ on the perpendicular bisector of a segment again and examine what we see:

We can't really see much from this. However, when we can't see much in a diagram, draw lines!

So, trying to apply the pythagorean theorem, we draw $\overline{XA}$ and $\overline{XB}$, and assign some lengths:

Aha! Now we can use the pythagorean theorem to get: $$a^2 + b^2 = c_1^2$$ $$a^2 + b^2 = c_2^2$$ But look at this repetition! We can use the transitive property to get: $$c_1^2 = c_2^2$$ Then we can take the square root: $$|c_1| = |c_2|$$ Since both $c_1$ and $c_2$ are side lengths, they're positive, so we have... $c_1 = c_2$!

Using this, we can define an alternate definition of perpendicular bisector:

The perpendicular bisector of a segment is the locus of points equidistant from the two endpoints of the segment.

Intuitively, this is true by pure symmetry.

In the eyes of a point $P$ on the perpendicular bisector, there is nothing that actually distinguishes points $A$ and $B$.

More formally, one may use congruent triangles to prove this result.

Denote the midpoint of $AB$ as $M$.

Clearly, $\triangle APM\cong\triangle BPM$ by SAS congruence.

Thus, $AP=BP$.

Therefore, we have proven that a point on the perpendicular bisector of a segment is equidistant from the two endpoints of the segment.