Question

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

Answer 1

$\angle DBC={70}^o,\angle BAC={30}^o$, then $\angle BCD=?$
$AB=BC,$ then $\angle ECD=?$
$\angle DAC$ and $\angle DBC$ are angles in same segment
$\therefore \angle DAC=\angle DBC={70}^o$
$\therefore \angle DAC={70}^o$

$\therefore \angle DAC+\angle BAC=70+30=100$
$ABCD$ is a cyclic quadrilateral
$\therefore$ Sum of opposite angles is ${180}^o$
$100+\angle DCB=180$
$[\because \angle DAC+\angle BAC=\angle DAB=100$]
$\angle DCB=180-100$
$\therefore \angle DCB=80$
$\angle DCB=\angle BCD=80$
$\therefore \angle BCD=80$
In $△ABC,AB=AC$
$\therefore \angle BAC=\angle BCA={30}^o$
$\angle BCA={30}^o$
$\angle ECD=\angle BCD-\angle BCA=80-30$
$\therefore \angle ECD={50}^o$