ABCD is a cyclic quadrilateral whose diagonals intersect at a point E

Solution:



For chord CD,

$\angle C B D=\angle C A D$ (Angles in the same segment)

$\angle C A D=70^{\circ}$

$\angle B A D=\angle B A C+\angle C A D=30^{\circ}+70^{\circ}=100^{\circ}$

$\angle B C D+\angle B A D=180^{\circ}$ (Opposite angles of a cyclic quadrilateral)

$\angle B C D+100^{\circ}=180^{\circ}$

$\angle B C D=80^{\circ}$

$\ln \triangle \mathrm{ABC}$.

$\mathrm{AB}=\mathrm{BC}$ (Given)

$\therefore \angle B C A=\angle C A B$ (Angles opposite to equal sides of a triangle)

$\Rightarrow \angle B C A=30^{\circ}$

We have, $\angle B C D=80^{\circ}$

$\Rightarrow \angle B C A+\angle A C D=80^{\circ}$

$30^{\circ}+\angle \mathrm{ACD}=80^{\circ}$

$\Rightarrow \angle \mathrm{ACD}=50^{\circ}$

$\Rightarrow \angle E C D=50^{\circ}$