Question

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

• A. 60^o
• B. 70^o
• C. 50^o
• D. 80^o

Answer 1

The correct option is C

50^o

$\angle CDB=\angle BAC={30}^{\circ }...\left(i\right)\text{[Angles in the same segment of a circle are equal]}\angle DBC={70}^{\circ }.....\left(ii\right)$
$In\Delta BCD,$
$\angle BCD+\angle DBC+\angle CDB={180}^{\circ }\text{[Sum of all the angles of a triangle is}{180}^{\circ }]$
$\angle BCD+{70}^{\circ }+{30}^{\circ }={180}^{\circ }\text{[Using (i) and (ii)]}$
$\angle BCD+{100}^{\circ }={180}^{\circ }$
$\angle BCD={180}^{\circ }-{100}^{\circ }$
$\angle BCD={80}^{\circ }.....\left(iii\right)$

$In\Delta ABC,AB=BC$
$\therefore \angle BCA=\angle BAC={30}^{\circ }\text{[Angle opposite to equal sides of a triangle are equal]}...\left(iv\right)\angle BAC={30}^{\circ }\left(given\right)$
$Now\angle BCD={80}^{\circ }$
$\Rightarrow \angle BCA+\angle ECD={80}^{\circ }$
$\Rightarrow {30}^{\circ }+\angle ECD={80}^{\circ }$
$\Rightarrow \angle ECD={80}^{\circ }-{30}^{\circ }$
$\Rightarrow \angle ECD={50}^{\circ }$