Congratulations to Peter Conlon, age 15, from Worth School in Sussex. This is Peter's solution.

Given five juxtaposing cyclic quadrilaterals $(C . Q .)$ , prove that the resulting quadrilateral, $EFGH$ in this case, is also a $C . Q .$

Taken for granted: In a $C . Q .$ , opposite angles add up to $180^{o}$ "
Let $∠$ DEF be $x^{o}$ , $∠$ FGB be $y^{o}$ and $∠$ HED be $z^{0}$

Because $CDEF$ is a $C . Q$ , $DCF$ = $180^{o} - x^{o}$
Because $CBFG$ is a $C . Q, BCF = 180^{o} - y^{o}$
Therefore $∠ DCB = (x + y)^{o} (360^{o}$ in a circle $)$

Because $ABCD$ is a $C . Q, DAB = 180^{o} - (x + y)^{o}$
Because $AHED$ is a $C . Q, DAH = 180^{o} - z^{o}$
Therefore $HAB = z^{o} + (x + y)^{o} = (z + x)^{o} + y^{o} (360^{o}$ in a circle $)$

Because $ABGH$ is a $C . Q$ , $BGH = 180^{o} - (z + x)^{o} - y^{o}$
Therefore $HGF = 180^{o} - (z + x)^{o} - y^{o} + y^{o} = 180^{o} - (z + x)^{o}$
But $HEF = (z + x)^{o}$ , and $HEF$ is the opposite angle to $HGF$ , and $HEF + HGF = 180^{o}$
Therefore $EFGH$ is a $C . Q$ .