Show that the opposite angles of a cyclic quadrilateral add up to \({180^ \circ }\) .

A quadrilateral ABCD is inscribed in a circle \(S\) . The four tangents to \(S\) at the vertices A, B, C and D form the edges of a quadrilateral EFGH. Given that EFGH is cyclic, show that AC and BD intersect at right angles.

Show that the locus of a point \({\rm{P'}}\) , which satisfies \(\overrightarrow {{\rm{QP'}}}  = k\overrightarrow {{\rm{QP}}} \) , is a circle \(C'\) , where k is a constant and \(0 < k < 1\) .

Show that the two tangents to \(C\) from Q are also tangents to \({\rm{C'}}\) .

recognition of relevant theorem     (M1)

eg \({\rm{D}}\hat {\rm{O}}{\rm{B}} = 2 \times {\rm{D}}\hat {\rm{A}}{\rm{B}}\)     A1

\({360^ \circ } - {\rm{D}}\hat {\rm{O}}{\rm{B}} = 2 \times {\rm{D}}\hat {\rm{C}}{\rm{B}}\)     A1

so \({\rm{D}}\hat {\rm{A}}{\rm{B}} + {\rm{D}}\hat {\rm{C}}{\rm{B}} = {\rm{18}}{0^ \circ }\)     AG

[3 marks]

diagram showing tangents EAF, FBG, GCH and HDE; diagonals cross at M.     M1

let \(x = {\rm{E}}\hat {\rm{D}}{\rm{A}} = {\rm{E}}\hat {\rm{A}}{\rm{D}}\) ; \(y = {\rm{B}}\hat {\rm{C}}{\rm{G}} = {\rm{C}}\hat {\rm{B}}{\rm{G}}\)     A1

\({\rm{D}}\hat {\rm{E}}{\rm{A}} + {\rm{H}}\hat {\rm{G}}{\rm{F}} = 180 - 2x + 180 - 2y = 360 - 2(x + y)\)     M1A1

\({\rm{C}}\hat {\rm{D}}{\rm{B}} = y\) and \({\rm{A}}\hat {\rm{C}}{\rm{D}} = x\) , as angles in alternate segments     M1A1

\({\rm{D}}\hat {\rm{M}}{\rm{C}} = 180 - (x + y) = \left( {\frac{1}{2}} \right)({\rm{D}}\hat {\rm{E}}{\rm{A}} + {\rm{H}}\hat {\rm{G}}{\rm{F}}) = {90^ \circ }\)     A1

so the diagonals cross at right angles     AG

[7 marks]

                                                                     M1

let \({\rm{O'}}\) be the point on OQ such \({\rm{O'P'}}\) is parallel to OP     A1

using similar triangles \({\rm{O'Q}} = k{\rm{OQ}}\) , so \({\rm{O'}}\) is a fixed point     M1A1

and \({\rm{O'P}} = k{\rm{OP}}\) which is constant     A1

so \({\rm{P'}}\) lies on a circle centre \({\rm{O'}}\)     R1

so the locus of \({\rm{P'}}\) is a circle     AG

[6 marks]

let one of the two tangents to \(C\) from Q touch \(C\) at T

the image of T lies on TQ     A1

and is a unique point \({{\rm{T'}}}\) on \({C'}\)     A1

so \({\rm{TT'}}\) is a common tangent and passes through Q     R1

the same is true for the other tangent     A1

so the two tangents to \(C\) from Q are also tangents to \({C'}\)     AG

[4 marks]

(a) Most candidates produced a valid answer, although a small minority used a circular argument.

(b) A few candidates went straight to the core of this question. However, many other candidates produced incoherent answers containing some true statements, some irrelevancies and some incorrect statements, based on a messy diagram.

(a) This was poorly answered. Many candidates failed to note that the points Q, P and its image were defined to be collinear, and tried to invoke the notion of the Apollonius Circle theory. Others tried a coordinate approach – in principle this could work, but is actually quite tricky without a sensible choice of axes and the origin.