Question

In fig 4.37 AB is a chord of a circle with centre O. Diameter CD is perpendicular to AB intersecting it at E. Show that $△$ABC is isosceles.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{CAE}\mathrm{and}△\mathrm{CBE}, \\ \mathrm{AE}=\mathrm{BE}\left(\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{c}\mathrm{hord}\right) \\ \angle \mathrm{AEC}=\angle \mathrm{BEC}\left(\mathrm{both}\mathrm{equal}\mathrm{to}90°\right) \\ \mathrm{CE}=\mathrm{CE}\left(\mathrm{common}\mathrm{side}\right) \\ \therefore △\mathrm{CAE}\cong △\mathrm{CBE}(\text{by}\mathrm{SAS}\mathrm{congruency}) \\ \Rightarrow \mathrm{CA}\hspace{0.17em}=\mathrm{CB}\left(\text{by}\mathrm{CPCT}\right) \\ \therefore △\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}. \end{gathered}$$