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 $\angle ECD$. [2 Marks]

Answer 1

Given
$\angle DBC={70}^{\circ },\angle BAC={30}^{\circ }$
For chord CD,
$\angle CBD=\angle CAD\text{(Angles in the same segment)}\angle CAD={70}^{\circ }$
$\angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$
$\angle BCD+\angle BAD={180}^{\circ }$ (Opposite angles of a cyclic quadrilateral)
$\angle BCD+{100}^{\circ }={180}^{\circ }$
$\angle BCD={80}^{\circ }$ [1 Mark]
In $\Delta ABC$,
AB = BC (Given)
$\Rightarrow \angle BCA=\angle CAB$ (Angles opposite to equal sides of a triangle)
$\therefore \angle BCA={30}^{\circ }$
We have, $\angle BCD={80}^{\circ }$
$\angle BCA+\angle ACD={80}^{\circ }$
${30}^{\circ }+\angle ACD={80}^{\circ }$
$\Rightarrow \angle ACD={50}^{\circ }$
$\therefore \angle ECD={50}^{\circ }$ [1 Mark]