Question

Sides AB and CD of a quadrilateral ABCD are produced as shown in the following figure. Show that $\angle x+\angle y=\angle a+\angle b$.

Answer 1

In quadrilateral ABCD
$$\begin{gathered} \angle CDA={180}^{\circ }-\angle y \\ \angle ABC={180}^{\circ }-\angle x \end{gathered}$$
Since, the sum of the angles of a quadrilateral is ${180}^{\circ }$, therefore
$$\begin{gathered} \angle CDA+\angle DAB+\angle ABC+\angle BCD={360}^{\circ } \\ \Rightarrow {180}^{\circ }-\angle y+\angle a+{180}^{\circ }-\angle x+\angle b={360}^{\circ } \\ \Rightarrow \angle a+\angle b=\angle x+\angle y \end{gathered}$$
Hence, $\angle a+\angle b=\angle x+\angle y$.