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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 .
AP : PD
Answer/Explanation
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]
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]
Question
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].
Answer/Explanation
Markscheme
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]
Question
The diagram below shows a quadrilateral ABCD and a straight line which intersects (AB), (BC), (CD), (DA) at the points P, Q, R, S respectively.
Using Menelaus’ theorem, show that \(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{CR}}}}{{{\rm{RD}}}} \times \frac{{{\rm{DS}}}}{{{\rm{SA}}}} = 1\) .
Answer/Explanation
Markscheme
join BD and let the transversal meet (BD) at T (A1)
apply Menelaus’ theorem to triangle ABD with transversal (RS) : M1
\(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BT}}}}{{{\rm{TD}}}} \times \frac{{{\rm{DS}}}}{{{\rm{SA}}}} = ( – )1\) A1
apply Menelaus’ theorem to triangle CBD with transversal (RS) : M1
\(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{CR}}}}{{{\rm{RD}}}} \times \frac{{{\rm{DT}}}}{{{\rm{TB}}}} = ( – )1\) A1
multiplying these two results,
\(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BT}}}}{{{\rm{TD}}}} \times \frac{{{\rm{DS}}}}{{{\rm{SA}}}} \times \frac{{{\rm{BQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{CR}}}}{{{\rm{RD}}}} \times \frac{{{\rm{DT}}}}{{{\rm{TB}}}} = 1\) M1A1
whence
\(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{CR}}}}{{{\rm{RD}}}} \times \frac{{{\rm{DS}}}}{{{\rm{SA}}}} = 1\) AG
Note: The question can also be solved by joining AC and letting the transversal meet (AC) at T. Menelaus’ Theorem then has to be applied to triangles ABC and ACD.
The relevant equations are \(\frac{{{\rm{AP}}}}{{{\rm{PB}}}} \times \frac{{{\rm{BQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{CT}}}}{{{\rm{TA}}}} = ( – )1\) and \(\frac{{{\rm{CT}}}}{{{\rm{TA}}}} \times \frac{{{\rm{AS}}}}{{{\rm{SD}}}} \times \frac{{{\rm{DR}}}}{{{\rm{RC}}}} = ( – )1\) .
[7 marks]
Question
The points A, B have coordinates \(( – 3,{\text{ }}0)\), \((5,{\text{ }}0)\) respectively. Consider the Apollonius circle \(C\) which is the locus of point P where
\[\frac{{{\text{AP}}}}{{{\text{BP}}}} = k{\text{ for }}k \ne 1.\]
Given that the centre of \(C\) has coordinates \((13,{\text{ }}0)\), find
(i) the value of \(k\);
(ii) the radius of \(C\);
(iii) the \(x\)-intercepts of \(C\).
Let M be any point on \(C\) and N be the \(x\)-intercept of \(C\) between A and B.
Prove that \({\rm{A\hat MN}} = {\rm{N\hat MB}}\).
Answer/Explanation
Markscheme
(i) let \((x,{\text{ }}y)\) be a point on \(C\)
then \({(x + 3)^2} + {y^2} = {k^2}\left( {{{(x – 5)}^2} + {y^2}} \right)\) M1A1A1
Note: Award M1 for form of an Apollonius circle, A1 for each side.
rearrange, for example,
\(({k^2} – 1){x^2} – (10{k^2} + 6)x + ({k^2} – 1){y^2} + 25{k^2} – 9 = 0\) A1
equate the \(x\)-coordinate of the centre as given by this equation to 13:
\(\frac{{5{k^2} + 3}}{{{k^2} – 1}} = 13\) M1A1
obtain \({k^2} = 2 \Rightarrow k = \sqrt 2 \) A1
(ii) METHOD 1
with this value of \(k\), the equation can be reduced to the form
\({(x – 13)^2} + {y^2} = 128\) M1A1
obtain the radius \(\sqrt {128} {\text{ }}\left( { = 8\sqrt 2 } \right)\) A1
METHOD 2
assuming N is the \(x\)-intercept of \(C\) between A and B
\(\frac{{{\text{AB}}}}{{{\text{BN}}}} = \frac{{16 – r}}{{r – 8}} = \sqrt 2 \) M1A1
\( \Rightarrow r = 8\sqrt 2 \) A1
Note: Accept answers given in terms of \(k\), if no value of \(k\) found in (a)(i).
(iii) \(x\)-intercepts are \(13 \pm 8\sqrt 2 \) A1
[11 marks]
because N lies on the circle it satisfies the Apollonius property
hence \({\text{AN}} = \sqrt 2 {\text{NB}}\) R1
but as \({\text{AM}} = \sqrt 2 {\text{MB}}\) R1
by the converse to the angle-bisector theorem R1
\({\rm{A\hat MN}} = {\rm{N\hat MB}}\) AG
[3 marks]
Question
The points P, Q and R, lie on the sides [AB], [AC] and [BC], respectively, of the triangle ABC. The lines (AR), (BQ) and (CP) are concurrent.
Use Ceva’s theorem to prove that [PQ] is parallel to [BC] if and only if R is the midpoint of [BC].
Answer/Explanation
Markscheme
suppose R is the midpoint of BC M1
Note: The first mark is for initiating a relevant discussion for “if” or “only if” by Ceva’s theorem.
\(\frac{{{\text{AP}}}}{{{\text{PB}}}} \times \frac{{{\text{BR}}}}{{{\text{RC}}}} \times \frac{{{\text{CQ}}}}{{{\text{QA}}}} = 1\) A1
\( \Rightarrow \frac{{{\text{AP}}}}{{{\text{PB}}}} = \frac{{{\text{AQ}}}}{{{\text{QC}}}}\) or equivalent A1
\( \Rightarrow \frac{{{\text{PB}}}}{{{\text{AP}}}} + 1 = \frac{{{\text{QC}}}}{{{\text{AQ}}}} + 1\) (M1)
\( \Rightarrow \frac{{{\text{AP}} + {\text{PB}}}}{{{\text{AP}}}} = \frac{{{\text{AQ}} + {\text{QC}}}}{{{\text{AQ}}}}\)
\( \Rightarrow \frac{{{\text{AB}}}}{{{\text{AP}}}} = \frac{{{\text{AC}}}}{{{\text{AQ}}}}\) A1
\( \Rightarrow \) triangles APQ and ABC are similar with common base angles R1
so PQ is parallel to BC AG
statement of the converse A1
the argument is reversible R1AG
[8 marks]
Question
Prove the internal angle bisector theorem, namely that the internal bisector of an angle of a triangle divides the side opposite the angle into segments proportional to the sides adjacent to the angle.
The bisector of the exterior angle \(\widehat A\) of the triangle ABC meets (BC) at P. The bisector of the interior angle \(\widehat B\) meets [AC] at Q. Given that (PQ) meets [AB] at R, use Menelaus’ theorem to prove that (CR) bisects the angle \({\rm{A}}\widehat {\rm{C}}{\rm{B}}\) .
Answer/Explanation
Markscheme
EITHER
let [AD] bisect A, draw a line through C parallel to (AD) meeting (AB) at E M1
then \({\rm{B}}\widehat {\rm{A}}{\rm{D}} = {\rm{A}}\widehat {\rm{E}}{\rm{C}}\) and \({\rm{D}}\widehat {\rm{A}}{\rm{C}} = {\rm{A}}\widehat {\rm{C}}{\rm{E}}\) A1
since \({\rm{B}}\widehat {\rm{A}}{\rm{D}} = {\rm{D}}\widehat {\rm{A}}{\rm{C}}\) it follows that \({\rm{A}}\widehat {\rm{E}}{\rm{C}} = {\rm{A}}\widehat {\rm{C}}{\rm{E}}\) A1
triangle AEC is therefore isosceles and \({\rm{AE}} = {\rm{AC}}\) A1
since triangles BAD and BEC are similar
\(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{AB}}}}{{{\rm{AE}}}} = \frac{{{\rm{AB}}}}{{{\rm{AC}}}}\) M1A1
OR
\(\frac{{{\rm{AB}}}}{{\sin \beta }} = \frac{{{\rm{BD}}}}{{\sin \alpha }}\) M1A1
\(\frac{{{\rm{AC}}}}{{\sin (180 – \beta )}} = \frac{{{\rm{DC}}}}{{\sin \alpha }}\) M1A1
\(\sin \beta = \sin (180 – \beta )\) R1
\( \Rightarrow \frac{{{\rm{AB}}}}{{{\rm{BD}}}} = \frac{{{\rm{AC}}}}{{{\rm{DC}}}}\)
\( \Rightarrow \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{AB}}}}{{{\rm{AC}}}}\) A1
[6 marks]
using the angle bisector theorem, M1
\(\frac{{{\rm{AQ}}}}{{{\rm{QC}}}} = \frac{{{\rm{AB}}}}{{{\rm{BC}}}}\) and \(\frac{{{\rm{BP}}}}{{{\rm{PC}}}} = \frac{{{\rm{AB}}}}{{{\rm{BC}}}}\) A1
using Menelaus’ theorem with (PR) as transversal to triangle ABC M1
\(\frac{{{\rm{BR}}}}{{{\rm{AR}}}} \times \frac{{{\rm{AQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{PC}}}}{{{\rm{BP}}}} = ( – )1\) A1
substituting the above results, M1
\(\frac{{{\rm{BR}}}}{{{\rm{AR}}}} \times \frac{{{\rm{AB}}}}{{{\rm{BC}}}} \times \frac{{{\rm{AC}}}}{{{\rm{AB}}}} = ( – )1\) A1
giving
\(\frac{{{\rm{BR}}}}{{{\rm{AR}}}} = \frac{{{\rm{BC}}}}{{{\rm{AC}}}}\) A1
[CR] therefore bisects angle C by (the converse to) the angle bisector theorem R1AG
[8 marks]