Example 10.1 A, B is a line segment and the line is perpendicular bisector to point P. Show that P is equidistant from A and B.

Given that $l$ is the perpendicular bisector of line segment AB, it means that $l$ intersects AB at its midpoint, let's call it M, and $l$ is perpendicular to AB.
Therefore, we have:

• AM = MB (M is the midpoint of AB)
• $\angle PMA = \angle PMB = 90^\circ$ (l is perpendicular to AB)

Let P be any point on the line $l$. We need to show that P is equidistant from A and B, which means we need to prove that PA = PB. Consider the triangles $\triangle PMA$ and $\triangle PMB$.

We can use the Side-Angle-Side (SAS) congruence criterion:

• PM = PM (Common side to both triangles)
• $\angle PMA = \angle PMB = 90^\circ$ (From Step 1, definition of perpendicular bisector)
• AM = MB (From Step 1, definition of perpendicular bisector) By SAS congruence criterion, $\triangle PMA \cong \triangle PMB$.

Since $\triangle PMA \cong \triangle PMB$, their corresponding parts are equal (CPCTC - Corresponding Parts of Congruent Triangles are Congruent).
Therefore, PA = PB.
This shows that any point P on the perpendicular bisector $l$ is equidistant from points A and B.

The final answer is $\text{PA = PB}$.