Question

question 4 of 5
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consider △pnm and △pno
since (overline{pn}congoverline{pn}) by the
congruence criterion. by
of congruence, then △pnm is congruent to △pno by the
△pnm across (overline{pn}) it maps onto △pno since △pnm can be mapped onto
△pno using only rigid transformations, this verifies the congruence criterion.

Explanation:

Step1: Identify common side

$\overline{PN}$ is common to $\triangle PNM$ and $\triangle PNO$, which is reflexive property of congruence (a side is congruent to itself).

Step2: Analyze congruence criterion

If we assume other conditions (not shown in full - perhaps right - angles or equal side - lengths from the markings), and since $\triangle PNM$ can be mapped onto $\triangle PNO$ across $\overline{PN}$ using rigid - transformations (reflection in this case), if two sides and the included angle are equal (SAS - Side - Angle - Side) or other valid congruence criteria are met, the triangles are congruent. But from the given $\overline{PN}\cong\overline{PN}$ by reflexive property.

Answer:

Since $\overline{PN}\cong\overline{PN}$ by the reflexive congruence criterion. By the Side - Angle - Side (SAS) or other appropriate congruence criterion (depending on other un - shown information), $\triangle PNM$ is congruent to $\triangle PNO$ by the rigid transformation of reflection across $\overline{PN}$.