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Date May 2009 Marks available 4 Reference code 09M.1.hl.TZ0.3
Level HL only Paper 1 Time zone TZ0
Command term Calculate Question number 3 Adapted from N/A

Question

Triangle ABC has points D, E and F on sides [BC], [CA] and [AB] respectively; [AD], [BE] and [CF] intersect at the point P. If 3BD = 2DC and CE = 4EA , calculate the ratios

AF : FB .

[4]
a.

AP : PD

[4]
b.

Markscheme

using Ceva's theorem,

\(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} \times \frac{{{\rm{AF}}}}{{{\rm{FB}}}} = 1\)     M1A1

\(\frac{2}{3} \times \frac{4}{1} \times \frac{{{\rm{AF}}}}{{{\rm{FB}}}} = 1\)     A1

\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{3}{8}\) or AF : FB \( = 3 : 8\)     A1

[4 marks]

a.

using Menelaus' theorem in triangle ACD with BPE as transversal

\(\frac{{{\rm{AE}}}}{{{\rm{EC}}}} \times \frac{{{\rm{CB}}}}{{{\rm{BD}}}} \times \frac{{{\rm{DP}}}}{{{\rm{PA}}}} = - 1\)     M1A1

\(\frac{1}{4} \times  - \frac{5}{2} \times \frac{{{\rm{DP}}}}{{{\rm{PA}}}} = - 1\)     A1

\(\frac{{{\rm{DP}}}}{{{\rm{PA}}}} = \frac{8}{5}\) or AP : PD = 5 : 8    A1

[4 marks]

b.

Examiners report

This proved difficult for many candidates and often the ratios and negative signs were "blurred".

a.

This proved difficult for many candidates and often the ratios and negative signs were "blurred".

b.

Syllabus sections

Topic 2 - Geometry » 2.4 » Angle bisector theorem; Apollonius’ circle theorem, Menelaus’ theorem; Ceva’s theorem; Ptolemy’s theorem for cyclic quadrilaterals.

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