Question

In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If $\angle DBC={55}^{\circ }and\angle BAC={45}^{\circ },$ find $\angle BCD$.

Answer 1

In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are joined. $\angle DBC={55}^{\circ }and\angle BAC={45}^{\circ }$

$\because \angle BAC$ and $\angle BDC$ are in the same segment.

$\therefore \angle BAC=\angle BDC={45}^{\circ }$

Now in $\Delta BCD$, $\angle DBC+\angle BDC+\angle BCD={180}^{\circ }$ (Sum of angles of a triangle)

$\Rightarrow {55}^{\circ }+{45}^{\circ }+\angle BCD={180}^{\circ }$

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

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

Hence, $\angle BCD={80}^{\circ }$