1. **Problem statement:** Given a circle with points P, Q, and T on it, and a tangent at T meeting the line PQ extended at R, prove: (a) $\frac{TQ}{TP} = \frac{TR}{PR}$ using similar triangles. (b) Show that $\frac{TP^2}{TQ^2} = \frac{PR}{QR}$. 2. **Step (a) - Using similar triangles:** - The tangent at T and the chord PQ form two triangles: $\triangle TQP$ and $\triangle TRP$. - Since RT is tangent at T, angle $\angle TPR = \angle TRP$ (alternate segment theorem). - Also, $\angle TQP = \angle TRP$ (angles subtended by the same chord). - Therefore, $\triangle TQP \sim \triangle TRP$ by AA similarity. 3. **From similarity:** $$\frac{TQ}{TP} = \frac{TR}{PR}$$ 4. **Step (b) - Show $\frac{TP^2}{TQ^2} = \frac{PR}{QR}$:** - From (a), $\frac{TQ}{TP} = \frac{TR}{PR}$ implies $\frac{TP}{TQ} = \frac{PR}{TR}$. - Square both sides: $$\left(\frac{TP}{TQ}\right)^2 = \left(\frac{PR}{TR}\right)^2$$ - By the tangent-secant theorem, $TR^2 = PR \cdot QR$. - Substitute $TR = \sqrt{PR \cdot QR}$: $$\left(\frac{PR}{TR}\right)^2 = \frac{PR^2}{TR^2} = \frac{PR^2}{PR \cdot QR} = \frac{PR}{QR}$$ 5. **Final result:** $$\frac{TP^2}{TQ^2} = \frac{PR}{QR}$$ This completes the proof.