In the figure, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is a tangent to t

Question

In the figure, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is a tangent to the circle at $A$. Show that $\Delta CAP$ is similar to $\Delta CBA$. Prove that $AC^2=CP \times CB$. State a triangle that is similar to $\Delta CAQ$. Hence, or otherwise, prove that $CP \times CB = CD \times CQ$.

Accepted Answer

(i) Proved: $\triangle CAP\sim\triangle CBA$ (ii) Proved: $AC^2=CP\times CB$ (iii) $\triangle CDA$ Hence: $CP\times CB=CD\times CQ$

Proofs in Plane Geometry - Similarity

Proofs in Plane Geometry - Similarity

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