Question

In the given figure, O is the centre of the circle and XOY is a diameter. If XZ is any other chord of the circle, show that XY > XZ.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{XOZ},\mathrm{we}\mathrm{have}: \\ \mathrm{OX}+\mathrm{OZ}>\mathrm{XZ} \\ \therefore \mathrm{OX}+\mathrm{OY}>\mathrm{XZ}(\because \mathrm{OX}=\mathrm{OY}=\mathrm{OZ}) \\ \mathrm{But}\mathrm{OX}+\mathrm{OY}=\mathrm{XY} \\ \mathrm{So},\mathrm{XY}>\mathrm{XZ} \end{gathered}$$
Hence Proved.