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, then find $\angle ECD.$

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${50}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is C ${50}^{\circ }$

Angles by the same chord in the same segment of a circle are equal.
$\therefore \angle BAC=\angle BDC={30}^{\circ }$
Given that $\angle DBC={70}^{\circ }.$
Using angle sum property in $△BCD,$ we have
$\angle BCD+\angle DBC+\angle CDB={180}^{\circ }.$
$\angle BCD+{70}^{\circ }+{30}^{\circ }={180}^{\circ }$
$⟹\angle BCD={180}^{\circ }–{70}^{\circ }–{30}^{\circ }={80}^{\circ }$
Now in $△ABC,$ AB = BC.
Since the angles opposite to equal sides of a triangle are equal, $\angle BCA=\angle BAC={30}^{\circ }.$
Now, $\angle BCA+\angle ECD=\angle BCD.$
$⟹{30}^{\circ }+\angle ECD={80}^{\circ }$
$⟹\angle ECD={80}^{\circ }–{30}^{\circ }={50}^{\circ }$