In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR. Fig. 6.29
Step-by-Step Solution
Key Concept: When a pair of corresponding sides of two triangles are parallel, the alternate interior angles formed with the transversals are equal. Hence two angles of one triangle are equal to two angles of the other triangle, establishing similarity by the AA (Angle‑Angle) criterion.
1. Identify the parallel sides\
Given PQ \parallel RS.\
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2. Use alternate interior angles\
- The line \(QO\) cuts the two parallel lines PQ and RS. Therefore\
$$\angle PQO = \angle RSO \quad\text{(alternate interior angles)}.$$\
- The line \(PO\) also cuts the two parallel lines PQ and RS. Hence\
$$\angle QPO = \angle SRO \quad\text{(alternate interior angles)}.$$\
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3. Third angle equality\
In any triangle, the sum of the interior angles is \(180^{\circ}\).\
For \(\triangle POQ\):\
$$\angle POQ = 180^{\circ} - (\angle PQO + \angle QPO).$$\
For \(\triangle SOR\):\
$$\angle SOR = 180^{\circ} - (\angle RSO + \angle SRO).$$\
Since \(\angle PQO = \angle RSO\) and \(\angle QPO = \angle SRO\), the remaining angles are also equal, i.e.,\
$$\angle POQ = \angle SOR.$$\
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4. Apply AA similarity criterion\
We have shown that two angles of \(\triangle POQ\) are respectively equal to two angles of \(\triangle SOR\):\
$$\angle PQO = \angle RSO, \quad \angle QPO = \angle SRO.$$\
Hence, by the AA criterion,\
$$\triangle POQ \sim \triangle SOR.$$\
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5. Conclusion\
Therefore, when PQ is parallel to RS, the triangles POQ and SOR are similar.
Correct Answer: Δ POQ is similar to Δ SOR (i.e., \(\triangle POQ \sim \triangle SOR\)).