Question

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

• A. ${60}^{\circ },{20}^{\circ }$
• B. ${70}^{\circ },{40}^{\circ }$
• C. ${80}^{\circ },{50}^{\circ }$
• D. ${80}^{\circ },{30}^{\circ }$

Answer 1

The correct option is C (${80}^{\circ },{50}^{\circ }$)

Since the angles in the same segment of a circle are equal,

$\angle CDB=\angle BAC={30}^{\circ }...\left(i\right)$
Given, $\angle DBC={70}^{\circ }...\left(ii\right)$
Using angle sum property in $\Delta BCD,$
$\angle BCD+\angle DBC+\angle CDB={180}^{\circ }.$
Using (i) and (ii),
$\angle BCD+{70}^{\circ }+{30}^{\circ }={180}^{\circ }.$
$\angle BCD={180}^{\circ }-{100}^{\circ }={80}^{\circ }.....\left(iii\right)$

$In\Delta ABC,AB=BC$
Since angles opposite to equal sides of a triangle are equal,
$\therefore \angle BCA=\angle BAC={30}^{\circ }...\left(iv\right)$
We have $\angle BCD={80}^{\circ }$
$\Rightarrow \angle BCA+\angle ECD={80}^{\circ }$
i.e., ${30}^{\circ }+\angle ECD={80}^{\circ }$
$\Rightarrow \angle ECD={80}^{\circ }-{30}^{\circ }={50}^{\circ }$