Question

In figure, $\overline{AB}$ is a diameter of the circle, $\overline{CD}$ is a chord equal to the radius of the circle. $\overrightarrow{AC}$ and $\overrightarrow{BD}$ when extended intersect at a point E. Prove that $\angle AEB={60}^o$.

Answer 1

$CD=$ radius $r$

$OC=OD=$ radius

$OCD$ is equilateral triangle

$\angle DCO=\angle COD=\angle ODC={60}^o$

$\angle ACB={90}^o$ (Angle in semicircle)

$\angle DOC=2\angle DBC$ (half angle)

$\angle DBC={30}^o$

$\angle ECB+\angle BCA={180}^o$ (linear pair)

$\angle ECB=180-90={90}^o$

In $△ECB$

$\angle CEB+\angle ECB+\angle CBE={180}^o$

$\angle CEB+90+30=180$

$\angle CEB={60}^o$