In the diagram, $PQS$ lies on a circle and $PCBQ$ lies on another circle. $PQ$ is a common cho

Question

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 $BC$ is parallel to $QS$.[3] Prove that triangle $PQM$ is similar to triangle $QSM$.[2] Hence show that $QM^2 = PM \times SM$.[2]

Accepted Answer

(a) $BC\parallel QS$ (proven) (b) $\triangle PQM\sim\triangle QSM$ (proven) (c) $QM^2=PM\times SM$ (shown)

2025 DSS PRELIMS P2 Q3

2025 DSS PRELIMS P2 Q3

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