Question

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

Answer 1

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,
$\therefore \angle BCA=\angle BAC={30}^{\circ }...\left(iv\right)$
$Wehave\angle BCD={80}^{\circ }$
$\Rightarrow \angle BCA+\angle ECD={80}^{\circ }$
i.e., ${30}^{\circ }+\angle ECD={80}^{\circ }$
$\Rightarrow \angle ECD={80}^{\circ }-{30}^{\circ }={50}^{\circ }$