Question

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

• A. ${50}^{\circ }$
• B. ${60}^{\circ }$
• C. ${40}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

The correct option is A

${50}^{\circ }$

For chord $CD,$
$\angle CBD=\angle CAD$ [Angles in the same segment]

$\Rightarrow \angle CAD={70}^{\circ }$
$\angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$
$\angle BCD+\angle BAD={180}^{\circ }$ [Opposite angles of a cyclic quadrilateral]
$\angle BCD+{100}^{\circ }={180}^{\circ }$
$\angle BCD={80}^{\circ }$
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, $\angle BCD={80}^{\circ }$
$\angle BCA+\angle ACD={80}^{\circ }$
${30}^{\circ }+\angle ACD={80}^{\circ }$
$\Rightarrow \angle ACD={50}^{\circ }$
$\therefore \angle ECD={50}^{\circ }$