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 $\angle ECD$.

[3 Marks]

Answer 1

Since the angles in the same segment of a circle are equal,
$\angle CDB=\angle DAC={70}^{\circ }...\left(i\right)$ $\left[1Mark\right]$
Given, $\angle BAC={30}^{\circ }...\left(ii\right)$
Using property of sum of opposite angles in a cyclic quadrilateral, we get
$\angle DAB+\angle BCD={180}^{\circ }.$
$\angle DAC+\angle CAB+\angle BCD={180}^{\circ }.$
Using (i) and (ii),
$\angle {70}^{\circ }+{30}^{\circ }+\angle BCD={180}^{\circ }.$
$\angle BCD={180}^{\circ }-{100}^{\circ }={80}^{\circ }.....\left(iii\right)$ $\left[1Mark\right]$

$In\Delta ABC,AB=BC$
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 }$$\left[1Mark\right]$