Question

If $AB=CD$ and $\angle AOB={50}^{\circ }$, then find the value of $\angle ODC$.

Answer 1

Given that AB = CD.
Since equal chords subtend equal angles at the centre,
$\angle AOB=\angle COD={50}^{\circ }$

Note that $△COD$ is an isosceles triangle as OC = OD = radius of the circle.
$⟹\angle ODC=\angle OCD=x\left(say\right)$
(Angles opposite to equal sides are equal)
Thus, using angle sum property in $△COD,$ we have
$\angle COD+\angle ODC+\angle OCD={180}^{\circ }$
$⟹{50}^{\circ }+2x={180}^{\circ }$
$⟹x=\frac{{130}^{\circ }}{2}={65}^{\circ }$
i.e., $\angle ODC={65}^{\circ }$