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. $\angle$s, cyclic quad.)} \\
x & = & 180^\circ – \theta
\end{array}$

Since $AB//ED$, we have

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

Consider cyclic quadrilateral $BCD$.

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

In $\Delta DEF$,

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