Question

In the figure, ABCD is a cyclic quadrilateral. Let AD, BC produced to meet at R; DC, AB produced to meet at S. Prove that $\hat{r}+\hat{s}={180}^o-2\hat{c}$.

Answer 1

In $△BCS$

$\angle CBS+\angle BCS+\angle BSC={180}^o\hat{c}+\hat{s}=180-\angle CBS\left(1\right)$
In $△ABR$

$\angle ARB+\angle RAB=\angle CBS$ (exterior angle)

$\hat{r}+\angle RAB=\angle CBS\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\hat{c}+\hat{s}=180-(\hat{r}+\angle RAB)\hat{s}+\hat{r}+\hat{c}=180-\left(\angle RAB\right)\left(3\right)$

$ABCD$ is cyclic quadrilateral

$\angle RAB+\angle DCB={180}^o\angle DCB+\angle BCS={180}^o\Rightarrow \angle RAB=\angle BCS=\hat{c}\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$

$\hat{s}+\hat{r}+\hat{c}=180-\hat{c}\hat{s}+\hat{r}=180-2\hat{c}$