IB DP Further Mathematics – 2.4 Angle bisector theorem; Apollonius’ circle theorem, Menelaus’ theorem; Ceva’s theorem; Ptolemy’s theorem for cyclic quadrilaterals HL Paper 1

 

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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 .

[4]
a.

AP : PD

[4]
b.
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]

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.

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 \(C\).

[11]
a.

Let M be any point on \(C\) and N be the \(x\)-intercept of \(C\) between A and B.

Prove that \({\rm{A\hat MN}} = {\rm{N\hat MB}}\).

[3]
b.
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]

a.

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]

b.

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.

[6]
a.

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 \({\rm{A}}\widehat {\rm{C}}{\rm{B}}\) .

[8]
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]

a.


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]

b.
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