Question

In the figure, $\angle BAD={40}^o$ then find $\angle BCD$.

Answer 1

In triangle $AOB$ and triangle $OCD$

$OA=OB=OC=OD$= radius

$\angle AOB=\angle COD$ .....[opposite angles]

By $SAS$ congruency both triangles are congruent

$\therefore \angle BAO=\angle OCD={40}^{\circ }$....[C.P.C.T.]