Question

use the given information to prove the following theorem. if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. we let p be any point on line l, but different from point n. given: $overline{pn}$ is the perpendicular bisector of $overline{qr}$ prove: $qp = rp$

Explanation:

Step1: Recall properties of perpendicular bisector

$\overline{PN}$ is perpendicular to $\overline{QR}$ and $N$ is the mid - point of $\overline{QR}$, so $QN=RN$ and $\angle QNP=\angle RNP = 90^{\circ}$.

Step2: Consider triangles $\triangle QNP$ and $\triangle RNP$

In $\triangle QNP$ and $\triangle RNP$, we have $QN = RN$ (from Step 1), $\angle QNP=\angle RNP$ (from Step 1), and $PN = PN$ (common side).

Step3: Apply Side - Angle - Side (SAS) congruence criterion

By the SAS congruence criterion, $\triangle QNP\cong\triangle RNP$.

Step4: Use congruent - triangle property

Since $\triangle QNP\cong\triangle RNP$, corresponding sides are equal. So, $QP = RP$.

Answer:

$QP = RP$