Question

$A,B,C,D,E$ in that order are points on a circle such that $AE$ is a diameter. Find $\angle CDE$ if $\angle ABC={120}^o$

Answer 1

From the figure, it can be seen that

$\measuredangle CDE=\measuredangle CDA+\measuredangle ADE$

$\measuredangle CDA={180}^0-{120}^0={60}^0$ (opposite angles of cyclic quadrilateral)

$\measuredangle ADE={90}^0$ (angle in semicircle by diameter)

$\measuredangle CDE={90}^0+{60}^0={150}^0$