Question

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

image of a triangle with vertices q, p, o, and points s on qp, r on qo, with \\(\overline{op} \parallel \overline{rs}\\)

answer attempt 1 out of 2
\\(\frac{pq}{op} = \frac{\boxed{}}{\boxed{}}\\)
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Explanation:

Step1: Identify Similar Triangles

Since \(\overline{OP} \parallel \overline{RS}\), by the Basic Proportionality Theorem (Thales' theorem) or AA (Angle - Angle) similarity criterion, \(\triangle QRS \sim \triangle QOP\) (because \(\angle Q\) is common and \(\angle QSR=\angle QPO\) (corresponding angles as \(RS\parallel OP\)) and \(\angle QRS = \angle QOP\) (corresponding angles)).

Step2: Apply Proportionality of Similar Triangles

For similar triangles \(\triangle QRS\) and \(\triangle QOP\), the ratios of corresponding sides are equal. The side corresponding to \(PQ\) in \(\triangle QRS\) is \(QS\), and the side corresponding to \(OP\) in \(\triangle QRS\) is \(RS\). Also, we can look at the other pair of sides. The ratio \(\frac{PQ}{OP}\) should be equal to \(\frac{QS}{RS}\) or \(\frac{QR}{QO}\). But from the given fraction \(\frac{PQ}{OP}\), we need to find the corresponding sides. Another way: In \(\triangle QOP\) and \(\triangle QRS\), \(\frac{PQ}{QS}=\frac{OP}{RS}\), which can be rearranged to \(\frac{PQ}{OP}=\frac{QS}{RS}\). Also, considering the sides from \(Q\) to \(R\) and \(Q\) to \(O\), we have \(\frac{PQ}{OP}=\frac{QR}{QO}\) (since \(PQ = QS\) and \(OP = RS\) in terms of the segments? Wait, no. Let's re - examine the triangle. The vertices are \(Q\), \(O\), \(P\) with \(S\) on \(QP\) and \(R\) on \(QO\), and \(RS\parallel OP\). So by the Basic Proportionality Theorem, \(\frac{QS}{SP}=\frac{QR}{RO}\), but we need the ratio \(\frac{PQ}{OP}\). Let's denote \(PQ\) as a segment on \(QP\) and \(OP\) as a segment. Wait, actually, the correct ratio from similar triangles \(\triangle QRS\sim\triangle QOP\) is \(\frac{QS}{QP}=\frac{RS}{OP}\), which can be rewritten as \(\frac{QP}{OP}=\frac{QS}{RS}\), but we have \(\frac{PQ}{OP}\) (note that \(PQ = QP\) since it's the same segment). So the ratio \(\frac{PQ}{OP}=\frac{QS}{RS}\) or \(\frac{QR}{QO}\). But the most appropriate one here, looking at the segments, is \(\frac{PQ}{OP}=\frac{QS}{RS}\) or \(\frac{QR}{QO}\). But if we consider the sides of the triangles, the correct corresponding sides for the ratio \(\frac{PQ}{OP}\) would be \(\frac{QS}{RS}\) (since \(PQ\) corresponds to \(QS\) and \(OP\) corresponds to \(RS\) in the similar triangles \(\triangle QRS\) and \(\triangle QOP\)). Alternatively, \(\frac{QR}{QO}\) (since \(QR\) corresponds to \(QP\) and \(QO\) corresponds to \(OP\)? No, let's do it properly. Let \(\triangle QRS\sim\triangle QOP\) (AA similarity: \(\angle Q\) is common, \(\angle QSR=\angle QPO\) as \(RS\parallel OP\)). Then the ratio of sides: \(\frac{QR}{QO}=\frac{QS}{QP}=\frac{RS}{OP}\). So \(\frac{QP}{OP}=\frac{QS}{RS}\), and since \(QP = PQ\) (segment), we have \(\frac{PQ}{OP}=\frac{QS}{RS}\). Also, \(\frac{PQ}{OP}=\frac{QR}{QO}\) (because \(\frac{QR}{QO}=\frac{QS}{QP}\) and \(\frac{QP}{OP}=\frac{QS}{RS}\), so cross - multiplying gives \(\frac{QR}{QO}=\frac{PQ}{OP}\)). But the most likely answer here, considering the segments, is \(\frac{PQ}{OP}=\frac{QR}{QO}\) or \(\frac{QS}{RS}\). But looking at the diagram, the segments are \(Q\) to \(S\) to \(P\) on \(QP\) and \(Q\) to \(R\) to \(O\) on \(QO\). So by the Basic Proportionality Theorem (Thales' theorem), \(\frac{QS}{SP}=\frac{QR}{RO}\), but we need \(\frac{PQ}{OP}\). Let's let \(PQ=x\), \(OP = y\), \(QS = m\), \(RS=n\). Since \(\triangle QRS\sim\triangle QOP\), \(\frac{m}{x}=\frac{n}{y}\), so \(\frac{x}{y}=\frac{m}{n}\), i.e., \(\frac{PQ}{OP}=\frac{QS}{RS}\). Also, \(\frac{PQ}{OP}=\frac{QR}{QO}\) because \(QR = QO - RO\) and \(PQ=QP - QS\), but from similarity, \(\frac{QR}{QO}=\frac{QS}{QP}=\frac{RS}{OP}\), so \(\frac{QP}{OP}=\frac{QS}{RS}\) and \(\frac{QR}{QO}=\frac{QS}{QP}\), so \(\frac{QP}{OP}=\frac{QR}{QO}\) (by transitivity: \(\frac{QP}{OP}=\frac{QS}{RS}\) and \(\frac{QR}{QO}=\frac{QS}{QP}\), multiply both sides of the second equation by \(\frac{QP}{OP}\): \(\frac{QR}{QO}\times\frac{QP}{OP}=\frac{QS}{RS}\), no, that's not right. Wait, let's start over. The triangles \(\triangle QRS\) and \(\triangle QOP\) are similar. So the ratio of \(QP\) (which is \(PQ\)) to \(QS\) is equal to the ratio of \(OP\) to \(RS\). So \(\frac{PQ}{QS}=\frac{OP}{RS}\), which can be rearranged to \(\frac{PQ}{OP}=\frac{QS}{RS}\). Also, the ratio of \(QO\) to \(QR\) is equal to the ratio of \(QP\) to \(QS\) (and \(OP\) to \(RS\)). So \(\frac{QO}{QR}=\frac{QP}{QS}=\frac{OP}{RS}\), which can be rearranged to \(\frac{QP}{OP}=\frac{QS}{RS}\) and \(\frac{QO}{QR}=\frac{QP}{QS}\), so \(\frac{QP}{OP}=\frac{QR}{QO}\) (by cross - multiplying \(\frac{QP}{QS}=\frac{OP}{RS}\) and \(\frac{QO}{QR}=\frac{QP}{QS}\), we get \(\frac{QO}{QR}=\frac{OP}{RS}\), then \(\frac{QP}{OP}=\frac{QS}{RS}\) and \(\frac{QO}{QR}=\frac{OP}{RS}\), so \(\frac{QP}{OP}=\frac{QS}{RS}=\frac{QR}{QO}\)). So the ratio \(\frac{PQ}{OP}\) should be equal to \(\frac{QR}{QO}\) (or \(\frac{QS}{RS}\)). But looking at the blanks, we need to put two segments. So the numerator and denominator of the right - hand side fraction. So \(\frac{PQ}{OP}=\frac{QR}{QO}\) (where \(QR\) is the segment from \(Q\) to \(R\) and \(QO\) is the segment from \(Q\) to \(O\)) or \(\frac{QS}{RS}\) (where \(QS\) is from \(Q\) to \(S\) and \(RS\) is from \(R\) to \(S\)). The most appropriate answer, considering the diagram, is \(\frac{PQ}{OP}=\frac{QR}{QO}\) (or \(\frac{QS}{RS}\)). But let's check the labels again. The points are \(Q\), \(S\), \(P\) on the top side, \(Q\), \(R\), \(O\) on the left side, and \(R\) to \(S\) is parallel to \(O\) to \(P\). So by the Basic Proportionality Theorem, \(\frac{QS}{SP}=\frac{QR}{RO}\), but we need \(\frac{PQ}{OP}\). Let's assume that \(PQ\) is the entire segment from \(P\) to \(Q\), \(OP\) is from \(O\) to \(P\), \(QS\) is from \(Q\) to \(S\), and \(RS\) is from \(R\) to \(S\). Since \(\triangle QRS\sim\triangle QOP\), \(\frac{PQ}{QS}=\frac{OP}{RS}\), so \(\frac{PQ}{OP}=\frac{QS}{RS}\). Also, \(\frac{PQ}{OP}=\frac{QR}{QO}\) because \(QO = QR+RO\) and \(PQ = QS + SP\), and from the similarity, \(\frac{QR}{QO}=\frac{QS}{PQ}\), so \(\frac{PQ}{OP}=\frac{QR}{QO}\) (by cross - multiplying \(\frac{QS}{PQ}=\frac{QR}{QO}\) and \(\frac{PQ}{OP}=\frac{QS}{RS}\), and \(\frac{QR}{QO}=\frac{RS}{OP}\), so \(\frac{PQ}{OP}=\frac{QR}{QO}\)). So the correct ratio is \(\frac{PQ}{OP}=\frac{QR}{QO}\) (or \(\frac{QS}{RS}\)). So we can fill in \(\frac{QR}{QO}\) or \(\frac{QS}{RS}\). The most likely one is \(\frac{QR}{QO}\) (or \(\frac{QS}{RS}\)). Let's take \(\frac{QR}{QO}\) as an example (or \(\frac{QS}{RS}\)).

Answer:

\(\frac{PQ}{OP}=\frac{\boldsymbol{QR}}{\boldsymbol{QO}}\) (or \(\frac{QS}{RS}\), but \(\frac{QR}{QO}\) is also correct. If we consider the segments, another correct answer is \(\frac{QS}{RS}\))