The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths $3$ cm and (\9\) cm respectively. E is a point on side [AB] such that AE is $3$ cm. Side [DE] is produced to meet the circumcircle of ABCD at point P. Use Ptolemy’s theorem to calculate the length of chord [AP].

construct diagonal [DB] and the chords [AP] and [PB]     M1

since ${\rm{D}}\hat {\rm{A}}{\rm{B}} = {90^ \circ }$ , [DB] is the diameter of the circle and ${\rm{DB}} = \sqrt {{9^2} + {3^2}}  = 3\sqrt {10}$     R1A1

triangle AED is a right-angled, isosceles triangle so ${\rm{DE}} = 3\sqrt 2$     R1A1

${\rm{A}}\hat {\rm{E}}{\rm{D}} = {\rm{P}}\hat {\rm{E}}{\rm{B}} = {45^ \circ }$     M1

$\Rightarrow {\rm{PB}} = {\rm{PE}} = 6\cos {\rm{P}}\hat {\rm{E}}{\rm{B}} = \frac{6}{{\sqrt 2 }} = 3\sqrt 2$     M1A1

using Ptolemy’s theorem in quadrilateral APBD

${\rm{PB}} \times {\rm{AD + AP}} \times {\rm{DB = DP}} \times {\rm{AB}}$     M1

$3\sqrt 2  \times 3 + {\rm{AP}} \times 3\sqrt {10}  = \left( {3\sqrt 2  + 3\sqrt 2 } \right) \times 9$     A1

${\rm{AP}} \times 3\sqrt {10}  = 54\sqrt 2  - 9\sqrt 2  = 45\sqrt 2$     A1

 ${\rm{AP}} = \frac{{45\sqrt 2 }}{{3\sqrt {10} }} = 3\sqrt 5$     A1

[12 marks]

The diagrams that some candidates drew were not always helpful to them and sometimes served to confuse what was required and make the problem harder than it was. Candidates were asked to use Ptolemy’s theorem but some ignored this request. Lengths of various segments were often written down without any evidence of where they came from.