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 z⁰

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.