Question

$ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\Delta DBC={70}^o$, $\Delta BAC$ is ${30}^o$, find $\angle BCD$. Further if $AB=BC$, find $\angle ECD$.

Answer 1

For chord $CD$,
$\angle CBD=\angle CAD$ ...Angles in same segment
$\angle CAD=70°$
$\angle BAD=\angle BAC+\angle CAD=30°+70°=100°$
$\angle BCD+\angle BAD=180°$ ...Opposite angles of a cyclic quadrilateral
$\Rightarrow \angle BCD+100°=180°$
$\Rightarrow \angle BCD=80°$
In $△ABC$
$AB=BC$ (given)
$\angle BCA=\angle CAB$ ...Angles opposite to equal sides of a triangle
$\angle BCA=30°$
Also, $\angle BCD=80°$
$\angle BCA+\angle ACD=80°$
$\Rightarrow 30°+\angle ACD=80°$
$\angle ACD=50°$
$\angle ECD=50°$