Question

In the figure, given below, AD = BC,

$\angle BAC={30}^{\circ }$ and

$\angle CBD={70}^{\circ }$. Find $\angle BCD$

Answer 1

In the figure, ABCD is a cyclic quadrilateral AC and BD are its diagonals $\angle BAC={30}^{\circ }$ and $\angle CBD={70}^{\circ }$

$\angle CAD=\angle CBD={70}^{\circ }$
(angles in same segment)

Similarly $\angle BAC=\angle BDC={30}^{\circ }$

$\therefore \angle BAD=\angle BAC+\angle CAD={30}^{\circ }+{70}^{\circ }={100}^{\circ }$

Now $\angle BCD+\angle BAD={180}^{\circ }$

(opposite angles of cyclic quad.)

$\Rightarrow \angle BCD+{100}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BCD={180}^{\circ }-{100}^{\circ }={80}^{\circ }$