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,
$\angle CBD=\angle CAD$ [Angle by a chord in the same segment are equal]

$\therefore \angle CAD=\angle CBD={70}^{\circ }$
$\angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$
$\angle BCD+\angle BAD={180}^{\circ }$ [Sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }$)
$\angle BCD+{100}^{\circ }={180}^{\circ }$
$\angle BCD={80}^{\circ }$

In $\Delta ABC$,
AB = BC (Given)
$\Rightarrow \angle BCA=\angle CAB$ (Angles opposite to equal sides of a triangle are equal)
$\therefore \angle BCA={30}^{\circ }$

We have, $\angle BCD={80}^{\circ }$
$\angle BCA+\angle ACD={80}^{\circ }$
${30}^{\circ }+\angle ACD={80}^{\circ }$
$\Rightarrow \angle ACD={50}^{\circ }$
$\therefore \angle ECD=\angle ACD={50}^{\circ }$