Question

given: \\(\overline{nh}\\) bisects \\(\overline{pj}\\) and \\(s\\) is the midpoint of \\(\overline{nh}\\)\
prove: \\(\overline{pn} \cong \overline{jh}\\)

Explanation:

Step1: Analyze Given Information

• $\overline{NH}$ bisects $\overline{PJ}$, so the intersection point (let's say the midpoint of $\overline{PJ}$ is $S$? Wait, no, $S$ is the midpoint of $\overline{NH}$. Wait, the diagram is a regular pentagon? So all the segments from the center $S$ to the vertices are equal? Wait, let's re - state the given:
• $\overline{NH}$ bisects $\overline{PJ}$: So if $\overline{NH}$ and $\overline{PJ}$ intersect at some point, and since it's a regular pentagon (from the diagram, the segments from $S$ to each vertex seem equal), let's assume $S$ is the center. So $\overline{NH}$ bisecting $\overline{PJ}$ means that the intersection point (which is $S$? Wait, the problem says $S$ is the midpoint of $\overline{NH}$. So $S$ is the midpoint of $\overline{NH}$, so $NS = SH$. Also, $\overline{NH}$ bisects $\overline{PJ}$, so the midpoint of $\overline{PJ}$ is the point where $\overline{NH}$ and $\overline{PJ}$ intersect, let's call that point $S$ (since in the diagram, all the lines meet at $S$). So $PS=SJ$.
• Also, in a regular pentagon, the distance from the center $S$ to each vertex is equal, so $SN = SJ=SP = SH$? Wait, no, let's use the given:
• Given $S$ is the midpoint of $\overline{NH}$, so $NS = HS$.
• Given $\overline{NH}$ bisects $\overline{PJ}$, so $PS = JS$.
• Also, $\angle PSN$ and $\angle JSH$: are they vertical angles? Yes, because $\overline{NH}$ and $\overline{PJ}$ intersect at $S$, so $\angle PSN=\angle JSH$ (vertical angles are equal).

Step2: Prove Triangles Congruent

Now, in $\triangle PSN$ and $\triangle JSH$:

• $PS = JS$ (from $\overline{NH}$ bisecting $\overline{PJ}$)
• $\angle PSN=\angle JSH$ (vertical angles)
• $NS = HS$ (from $S$ being the midpoint of $\overline{NH}$)

By the Side - Angle - Side (SAS) congruence criterion, $\triangle PSN\cong\triangle JSH$.

Step3: Conclude the Segments are Congruent

Since $\triangle PSN\cong\triangle JSH$, the corresponding sides $\overline{PN}$ and $\overline{JH}$ are congruent (corresponding parts of congruent triangles are congruent, CPCTC).

Answer:

By showing $\triangle PSN\cong\triangle JSH$ (SAS: $PS = JS$, $\angle PSN=\angle JSH$, $NS = HS$) and using CPCTC, we prove $\overline{PN}\cong\overline{JH}$.