Question

Let two circles intersect each other at points A and D. Let the diameter AB intersect the circle with centre P at point N and the diameter AC intersect the circle at point M with centre Q. Then prove that AC. AM = AB. AN

Answer 1

$$\begin{gathered} \mathrm{In}∆\mathrm{ACN}\mathrm{and}∆\mathrm{ABM},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{MAB}=\angle \mathrm{NAC}(\mathrm{Vertically}\mathrm{opposite}\mathrm{angle}) \\ \angle \mathrm{AMB}=\angle \mathrm{ANC}(\mathrm{Angle}\mathrm{in}\mathrm{the}\mathrm{semicircle}\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{angle}) \\ \therefore ∆\mathrm{ACN}~∆\mathrm{ABM}(\mathrm{BY}\mathrm{AA}\mathrm{similarity}) \\ \mathrm{i}.\mathrm{e}.,\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AM}} \\ \Rightarrow \mathrm{AC}\times \mathrm{AM}=\mathrm{AB}\times \mathrm{AN} \end{gathered}$$
Hence, proved.