2018-I-08

Ans: $x={180}^{\circ }–\theta$, $y=2\theta –{180}^{\circ }$
Consider cyclic quadrilateral $ABCD$.

$\begin{array}{rcll}x+\theta & =& {180}^{\circ }& \text{(opp. }\mathrm{\angle }\text{s, cyclic quad.)}\\ x& =& {180}^{\circ }–\theta \end{array}$

Since $AB//ED$, we have

$\begin{array}{rcll}\mathrm{\angle }ADE& =& x& \text{(alt. }\mathrm{\angle }\text{s, }AB//ED\text{)}\\ \mathrm{\angle }ADE& =& {180}^{\circ }–\theta \end{array}$

Consider cyclic quadrilateral $BCD$.

$\begin{array}{rcll}\mathrm{\angle }BED+\theta & =& {180}^{\circ }& \text{(opp. }\mathrm{\angle }\text{s, cyclic quad.)}\\ \mathrm{\angle }BED& =& {180}^{\circ }–\theta \end{array}$

In $\mathrm{\Delta }DEF$,

$\begin{array}{rcll}y& =& {180}^{\circ }–\mathrm{\angle }FED–\mathrm{\angle }FDE& \text{(}\mathrm{\angle }\text{ sum of }\mathrm{\Delta }\text{)}\\ y& =& {180}^{\circ }–({180}^{\circ }-\theta )–({180}^{\circ }–\theta )\\ y& =& 2\theta –{180}^{\circ }\end{array}$