<p>In the diagram, \(PQS\) lies on a circle and \(PCBQ\) lies on another circle. \(PQ\) is a common chord of the two circles. \(BPS\) is a straight line. \(QC\) is a tangent to the circle \(PQS\) at \(Q\). \(QC\) and \(BP\) intersect at \(M\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1772944780952-s8kos0n5k9i.png" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a-4"><div class="list-item-content"><span class="list-item-text">Prove that&nbsp;\(BC\) is parallel to \(QS\).</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b-4"><div class="list-item-content"><span class="list-item-text">Prove that triangle \(PQM\) is similar to triangle \(QSM\).</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c-4"><div class="list-item-content"><span class="list-item-text">Hence show that&nbsp;\(QM^2 = PM \times SM\).</span><span class="inline-marks">[2]</span></div></li></ol>

In the diagram, \(PQS\) lies on a circle and \(PCBQ\) lies on another circle. \(PQ\) is a common chord of the two circles. \(BPS\) is a straight line. \(QC\) is a tangent to the circle \(PQS\) at \(Q\). \(QC\) and \(BP\) intersect at \(M\).

Prove that triangle \(PQM\) is similar to triangle \(QSM\).

2025 DSS PRELIMS P2 Q3

2025 DSS PRELIMS P2 Q3