Question

given \\(\overline{lm} \parallel \overline{op}\\). enter segments in the blanks provided that would result in a true equation.

image of a triangle with points n, p, m, o, l. n is the top vertex, l is the bottom vertex, m is the right vertex. o is on nl, p is on nm. \\(\overline{lm} \parallel \overline{op}\\).

answer attempt 1 out of 2
\\(\dfrac{op}{on} = \dfrac{\boxed{\quad}}{\boxed{\quad}}\\)
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Explanation:

Step1: Identify Similar Triangles

Since $\overline{LM} \parallel \overline{OP}$, by the Basic Proportionality Theorem (Thales' theorem), $\triangle NOP \sim \triangle NLM$ (similar triangles).

Step2: Corresponding Sides Ratio

In similar triangles, the ratios of corresponding sides are equal. For $\triangle NOP$ and $\triangle NLM$, the corresponding sides are $OP$ and $LM$, $ON$ and $NL$. Wait, no, let's check the vertices. $N$ is common, $O$ is on $NL$, $P$ is on $NM$. So $\triangle NOP \sim \triangle NLM$ (AA similarity, as $\angle N$ is common and $\angle NOP = \angle NLM$ because $OP \parallel LM$). So the ratio of $OP$ to $LM$ should be equal to the ratio of $ON$ to $NL$? Wait, no, wait the given fraction is $\frac{OP}{ON}$. Let's re - examine.

Wait, maybe I made a mistake. Let's look at the segments. $OP \parallel LM$, so $\triangle NOP$ and $\triangle NLM$ are similar. So $\frac{OP}{LM}=\frac{ON}{NL}=\frac{NP}{NM}$. But the given is $\frac{OP}{ON}=\frac{?}{?}$. Wait, maybe the corresponding sides are $OP$ and $LM$, $ON$ and $NL$? No, wait, let's list the sides:

In $\triangle NOP$: sides are $ON$, $OP$, $NP$.

In $\triangle NLM$: sides are $NL$, $LM$, $NM$.

Since they are similar, $\frac{OP}{LM}=\frac{ON}{NL}=\frac{NP}{NM}$. But we need $\frac{OP}{ON}=\frac{?}{?}$. Let's rearrange the ratio. From $\frac{OP}{LM}=\frac{ON}{NL}$, we can cross - multiply to get $OP\times NL = LM\times ON$, then $\frac{OP}{ON}=\frac{LM}{NL}$. Wait, no, maybe I mixed up the correspondence.

Wait, the correct correspondence: $\angle N$ is common, $\angle NPO=\angle NML$ (corresponding angles, since $OP \parallel LM$), so $\triangle NOP \sim \triangle NLM$ with correspondence $N
ightarrow N$, $O
ightarrow L$, $P
ightarrow M$. Oh! That's the mistake. So $O$ corresponds to $L$, $P$ corresponds to $M$. So the sides: $NO$ corresponds to $NL$, $OP$ corresponds to $LM$, $NP$ corresponds to $NM$. So $\frac{OP}{LM}=\frac{NO}{NL}=\frac{NP}{NM}$.

But the given fraction is $\frac{OP}{ON}$. Let's write $ON$ as a segment on $NL$. Let $NL = ON + OL$. But maybe the correct ratio is $\frac{OP}{ON}=\frac{LM}{NL}$. Wait, no, let's do it properly.

If $\triangle NOP \sim \triangle NLM$ (with $O
ightarrow L$, $P
ightarrow M$), then $\frac{OP}{LM}=\frac{ON}{NL}$. So cross - multiplying, $OP\times NL=LM\times ON$, then $\frac{OP}{ON}=\frac{LM}{NL}$. But maybe the problem expects $\frac{OP}{ON}=\frac{LM}{NL}$? Wait, no, maybe I got the correspondence wrong.

Wait, let's start over. Let's label the triangle: $N$ is the top vertex, $L$ and $M$ are the bottom vertices, $O$ is on $NL$, $P$ is on $NM$, and $OP \parallel LM$. So by the Basic Proportionality Theorem (Thales' theorem), which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. Wait, Thales' theorem is about a line parallel to one side intersecting the other two sides, but here $OP$ is parallel to $LM$, and $OP$ is a line from $O$ on $NL$ to $P$ on $NM$, and $LM$ is the base. So actually, $\triangle NOP$ and $\triangle NLM$ are similar (AA similarity: $\angle N$ is common, $\angle NOP=\angle NLM$ because $OP \parallel LM$). So the ratio of corresponding sides: $\frac{OP}{LM}=\frac{ON}{NL}=\frac{NP}{NM}$.

But the given fraction is $\frac{OP}{ON}$. Let's solve for $\frac{OP}{ON}$. From $\frac{OP}{LM}=\frac{ON}{NL}$, we can rewrite it as $\frac{OP}{ON}=\frac{LM}{NL}$.

Wait, but maybe the answer is $\frac{LM}{NL}$? Wait, no, maybe I made a mistake in correspondence. Let's check the segments again. The triangle has vertices $N$, $L$, $M$. $O$ is on $NL$, $P$ is on $NM$, and $OP \parallel LM$. So by the Basic Proportionality Theorem (Thales' theorem), $\frac{NO}{NL}=\frac{NP}{NM}$. But we need to find $\frac{OP}{ON}=\frac{?}{?}$. Since $\triangle NOP \sim \triangle NLM$ (AA similarity), the ratio of $OP$ to $LM$ is equal to the ratio of $ON$ to $NL$. So $\frac{OP}{LM}=\frac{ON}{NL}$, which can be rearranged to $\frac{OP}{ON}=\frac{LM}{NL}$.

Alternatively, maybe the correct ratio is $\frac{LM}{NL}$. Wait, but let's think of the sides. If we consider the similar triangles, the side corresponding to $OP$ is $LM$, and the side corresponding to $ON$ is $NL$. So $\frac{OP}{ON}=\frac{LM}{NL}$.

Answer:

$\frac{LM}{NL}$ (or in the blanks, the top blank is $LM$ and the bottom blank is $NL$)