Question
The diagram shows triangle ABC together with its inscribed circle. Show that [AD], [BE] and [CF] are concurrent.
PQRS is a parallelogram and T is a point inside the parallelogram such that the sum of \({\rm{P}}\hat {\rm{T}}{\rm{Q}}\) and \({\rm{R}}\hat {\rm{T}}{\rm{S}}\) is \({180^ \circ }\) . Show that \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\) .
Answer/Explanation
Markscheme
Since the lengths of the two tangents from a point to a circle are equal (M1)
\({\rm{AF = AE, BF = BD, CD = CE}}\) A1
Consider
\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = 1\) (signed lengths are not relevant here) M2A2
It follows by the converse to Ceva’s Theorem that [AD], [BE], [CF] are concurrent. R2
[8 marks]
Draw the \(\Delta {\rm{PQU}}\) congruent to \(\Delta {\rm{SRT}}\) (or translate \(\Delta {\rm{SRT}}\) to form \(\Delta {\rm{PQU}}\) ). R2
Since \({\rm{P}}\hat {\rm{U}}{\rm{Q}} = {\rm{S}}\hat {\rm{T}}{\rm{R}}\) , and \({\rm{S}}\hat {\rm{T}}{\rm{R + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\) it follows that R2
\({\rm{P}}\hat {\rm{U}}{\rm{Q + P}}\hat {\rm{T}}{\rm{Q}} = {180^ \circ }\) R1
The quadrilateral PUQT is therefore cyclic R2
Using Ptolemy’s Theorem, M2
\({\rm{UQ}} \times {\rm{PT}} + {\rm{PU}} \times {\rm{QT}} = {\rm{PQ}} \times {\rm{UT}}\) A2
Since \({\rm{UQ = TR}}\) , \({\rm{PU = ST}}\) and \({\rm{UT = QR}}\) R2
Then \({\rm{TP}} \times {\rm{TR}} + {\rm{ST}} \times {\rm{TQ}} = {\rm{PQ}} \times {\rm{QR}}\) AG
[13 marks]
Question
The diagram shows the line \(l\) meeting the sides of the triangle ABC at the points D, E and F. The perpendiculars to \(l\) from A, B and C meet \(l\) at G, H and I.
(i) State why \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}}\) .
(ii) Hence prove Menelaus’ theorem for the triangle ABC.
(iii) State and prove the converse of Menelaus’ theorem.
A straight line meets the sides (PQ), (QR), (RS), (SP) of a quadrilateral PQRS at the points U, V, W, X respectively. Use Menelaus’ theorem to show that\[\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1.\]
Answer/Explanation
Markscheme
(i) Because the triangles AGF and BHF are similar. R1
(ii) It follows (by cyclic rotation or considering similar triangles) that
\(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{BH}}}}{{{\rm{IC}}}}\) A1
and \(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\) A1
Multiplying these three results gives Menelaus’ Theorem, i.e.
\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}} \times \frac{{{\rm{BH}}}}{{{\rm{IC}}}} \times \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\) M1A1
\( = \frac{{{\rm{AG}}}}{{{\rm{GA}}}} \times \frac{{{\rm{BH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{CI}}}}{{{\rm{IC}}}} = – 1\) M1A1
(iii) The converse states that if D, E, F are points on the sides (BC), (CA), (AB) of a triangle such that
\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = – 1\)
then D, E, F are collinear. A1
To prove this result, let D, E, F′ be collinear points on the three sides so that, using the above theorem, M1
\(\frac{{{\rm{AF’}}}}{{{\rm{F’B}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = – 1\) A1
Since \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = – 1\) M1
\(\frac{{{\rm{AF’}}}}{{{\rm{F’B}}}} = \frac{{{\rm{AF}}}}{{{\rm{FB}}}}\) A1
and \({\rm{F = F’}}\) which proves the converse. R1
[13 marks]
Draw the diagonal PR and let it cut the line at the point Y. M1
Apply Menelaus’ Theorem to the triangle PQR. Then,
\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} = – 1\) M1A1
Now apply the theorem to triangle PRS.
\(\frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = – 1\) A1
\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = – 1 \times – 1\) M1
\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YR}}}} = 1\) A1
\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times ( – 1) \times ( – 1) = 1\) (M1)
\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1\) AG
[7 marks]
Examiners report
[N/A]
[N/A]
Question
Prove that the interior bisectors of two of the angles of a non-isosceles triangle and the exterior bisector of the third angle, meet the sides of the triangle in three collinear points.
An equilateral triangle QRT is inscribed in a circle. If S is any point on the arc QR of the circle,
(i) prove that \({\rm{ST}} = {\rm{SQ}} + {\rm{SR}}\) ;
(ii) show that triangle RST is similar to triangle PSQ where P is the intersection of [TS] and [QR];
(iii) using your results from parts (i) and (ii) deduce that \(\frac{1}{{{\rm{SP}}}} = \frac{1}{{{\rm{SQ}}}} + \frac{1}{{{\rm{SR}}}}\) .
Perpendiculars are drawn from a point P on the circumcircle of triangle LMN to the three sides. The perpendiculars meet the sides [LM], [MN] and [LN] at the points E, F and G respectively.
Prove that \({\rm{PL}} \times {\rm{PF}} = {\rm{PM}} \times {\rm{PG}}\) .
Answer/Explanation
Markscheme
triangle ABC has interior angle bisectors AH, BG and exterior angle bisector CK
using the angle bisector theorem, M1
\(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} = \frac{{{\rm{CA}}}}{{{\rm{AB}}}}\) , \(\frac{{{\rm{AG}}}}{{{\rm{GC}}}} = \frac{{{\rm{AB}}}}{{{\rm{CB}}}}\) , \(\frac{{{\rm{BK}}}}{{{\rm{AK}}}} = \frac{{{\rm{CB}}}}{{{\rm{CA}}}}\) A2
hence, \(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{AG}}}}{{{\rm{GC}}}} \times \frac{{{\rm{BK}}}}{{{\rm{AK}}}} = \frac{{{\rm{CA}}}}{{{\rm{AB}}}} \times \frac{{{\rm{AB}}}}{{{\rm{CB}}}} \times \frac{{{\rm{CB}}}}{{{\rm{CA}}}} = 1\) M1A1
but \(\frac{{{\rm{BK}}}}{{{\rm{AK}}}} = – \frac{{{\rm{BK}}}}{{{\rm{KA}}}}\) R1
so, \(\frac{{{\rm{CH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{AG}}}}{{{\rm{GC}}}} \times \frac{{{\rm{BK}}}}{{{\rm{AK}}}} = – 1\) A1
hence, by converse Menelaus’ theorem, G, H and K are collinear R1
[8 marks]
(i)
\({\rm{ST}} \bullet {\rm{QR}} = {\rm{SQ}} \bullet {\rm{RT}} + {\rm{SR}} \bullet {\rm{QT}}\) M1A1
but \({\rm{QT = QR = RT}}\) R1
hence \({\rm{ST = SQ + SR}}\) AG
(ii) \(\angle {\rm{STR = }}\angle {\rm{SQR}}\) R1
\(\angle {\rm{QST = }}\angle {\rm{QRT = }}{60^ \circ }\) A1
\(\angle {\rm{RST = }}\angle {\rm{RQT = }}{60^ \circ }\) A1
hence \(\angle {\rm{QST = }}\angle {\rm{RST}}\) and \(\Delta {\rm{RST}} \sim \Delta {\rm{PSQ}}\) R1
(iii) \(\frac{{{\rm{ST}}}}{{{\rm{SQ}}}} = \frac{{{\rm{SR}}}}{{{\rm{SP}}}} \to {\rm{ST}} \times {\rm{SP}} = {\rm{SR}} \times {\rm{SQ}}\) M1A1
but \({\rm{ST = (SQ + SR)}}\)
\({\rm{(SQ}} + {\rm{SR) \times SP}} = {\rm{SR}} \times {\rm{SQ}}\) A1
\({\rm{SQ}} \times {\rm{SP}} + {\rm{SR}} \times {\rm{SP}} = {\rm{SR}} \times {\rm{SQ}}\)
hence \(\frac{1}{{{\rm{SP}}}} = \frac{1}{{{\rm{SQ}}}} + \frac{1}{{{\rm{SR}}}}\) AG
[10 marks]
sine \(m\angle {\rm{PEM}} \cong m\angle {\rm{PFM}} \cong {90^ \circ }\)
then quadrilateral PFEM is cyclic M1A1
so \(m\angle {\rm{PME}} \cong m\angle {\rm{PFG}}\)
since \(m\angle {\rm{PGL}} \cong m\angle {\rm{PEL}} \cong {90^ \circ }\)
then quadrilateral PGLE is cyclic
so \(m\angle {\rm{PGE}} \cong m\angle {\rm{PLE}}\) R1A1
now E, F and G are collinear since they are on the Simson line of \(\Delta {\rm{LMN}}\) R1
hence \(\Delta {\rm{PFG}} \sim \Delta {\rm{PML}}\) A1
so \(\frac{{{\rm{PL}}}}{{{\rm{PM}}}} = \frac{{{\rm{PG}}}}{{{\rm{PF}}}} \Rightarrow {\rm{PL}} \times {\rm{PF}} = {\rm{PM}} \times {\rm{PG}}\) R1AG
[8 marks]
Examiners report
As in paper 1 there is a sad lack of knowledge of geometry although some good solutions were seen and at least one school is using techniques very successfully that are not mentioned in the program.
Part (a) was well done.
Few clear solutions to part (b) were seen.
Question
The points D, E, F lie on the sides [BC], [CA], [AB] of the triangle ABC and [AD], [BE], [CF] intersect at the point G. You are given that CD \( = 2\)BD and AG \( = 2\)GD .
By considering (BE) as a transversal to the triangle ACD, show that
\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{3}{2}\) .
Determine the ratios
(i) \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}}\) ;
(ii) \(\frac{{{\rm{BG}}}}{{{\rm{GE}}}}\) .
The diagram shows a hexagon ABCDEF inscribed in a circle. All the sides of the hexagon are equal in length. The point P lies on the minor arc AB of the circle. Using Ptolemy’s theorem, show that\[{\rm{PE}} + {\rm{PD}} = {\rm{PA}} + {\rm{PB}} + {\rm{PC}} + {\rm{PF}}\]
Answer/Explanation
Markscheme
using Menelaus’ theorem in \(\Delta {\rm{ACD}}\) ,
\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{{{\rm{AG}}}}{{{\rm{GD}}}} \bullet \frac{{{\rm{DB}}}}{{{\rm{BC}}}} = – 1\) M1
\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{2}{1} \bullet \frac{1}{3} = 1\) A1
\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{3}{2}\) AG
[2 marks]
(i) using Ceva’s theorem in \(\Delta {\rm{ABC}}\) ,
\(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} \bullet \frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = 1\) M1
\(\frac{3}{2} \bullet \frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{1}{2} = 1\) A1
\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{4}{3}\) A1
(ii) using Menelaus’ theorem in \(\Delta {\rm{ABE}}\) , with traversal (FC), M1
\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \bullet \frac{{{\rm{BG}}}}{{{\rm{GE}}}} \bullet \frac{{{\rm{EC}}}}{{{\rm{CA}}}} = – 1\) A1
\(\frac{4}{3} \bullet \frac{{{\rm{BG}}}}{{{\rm{GE}}}} \bullet \frac{3}{5} = 1\) A1
\(\frac{{{\rm{BG}}}}{{{\rm{GE}}}} = \frac{5}{4}\) A1
[7 marks]
using Ptolemy’s theorem in PAEC, M1
\({\rm{PA}} \bullet {\rm{EC}} + {\rm{AE}} \bullet {\rm{PC = PE}} \bullet {\rm{AC}}\) A1
since \({\rm{EC = AE = AC}}\) , M1
\({\rm{PE = PA + PC}}\) A1
similarly for PBDF, M1
\({\rm{PB \bullet DF + BD \bullet PF = PD \bullet BF}}\) (A1)
\({\rm{PD = PB + PF}}\) A1
adding these results,
\({\rm{PE + PD = PA + PB + PC + PF}}\) AG
[7 marks]
Question
ABCD is a quadrilateral. (AD) and (BC) intersect at F and (AB) and (CD) intersect at H. (DB) and (CA) intersect (FH) at G and E respectively. This is shown in the diagram below.
Prove that \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = – \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\) .
Answer/Explanation
Markscheme
in \(\Delta {\rm{HFD}}\) , [HA], [FC] and [DG] are concurrent at B M1
so, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = 1\) by Ceva’s theorem A1R1
in \(\Delta {\rm{HFD}}\) , with CAE as transversal, M1
\(\frac{{{\rm{HE}}}}{{{\rm{EF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = – 1\) by Menelaus’ theorem A1R1
therefore, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = – \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\) AG
[6 marks]
Question
The diagram above shows a point \({\text{O}}\) inside a triangle \({\text{ABC}}\). The lines \({\text{(AO), (BO), (CO)}}\) meet the lines \({\text{(BC), (CA), (AB)}}\) at the points \({\text{D, E, F}}\) respectively. The lines \({\text{(EF), (BC)}}\) meet at the point \({\text{G}}\).
(a) Show that, with the usual convention for the signs of lengths in a triangle, \(\frac{{{\text{BD}}}}{{{\text{DC}}}} = – \frac{{{\text{BG}}}}{{{\text{GC}}}}\).
(b) The lines \({\text{(FD), (CA)}}\) meet at the point \({\text{H}}\) and the lines \({\text{(DE), (AB)}}\) meet at the point \({\text{I}}\). Show that the points \({\text{G, H, I}}\) are collinear.
Answer/Explanation
Markscheme
(a) applying Ceva’s theorem to triangle \({\text{ABC}}\),
\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{AE}}}}{{{\text{EC}}}} \times \frac{{{\text{BF}}}}{{{\text{FA}}}} = 1\) M1A1
applying Menelaus’ theorem to triangle \({\text{ABC}}\) with transversal \({\text{(GFE)}}\),
\(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} – = 1\) M1A1
multiplying the two equations, M1
\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{BG}}}}{{{\text{GC}}}} = – 1\) A1
so that \(\frac{{{\text{BD}}}}{{{\text{DC}}}} = – \frac{{{\text{BG}}}}{{{\text{GC}}}}\) AG
[6 marks]
(b) similarly
\(\frac{{{\text{CE}}}}{{{\text{EA}}}} = – \frac{{{\text{CH}}}}{{{\text{HA}}}}\) M1A1
and \(\frac{{{\text{AF}}}}{{{\text{FB}}}} = – \frac{{{\text{AI}}}}{{{\text{IB}}}}\) A1
multiplying the three results,
\(\frac{{{\text{BD}}}}{{{\text{DC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} = – \frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}}\) M1A1
by Ceva’s theorem, as shown previously, the left hand side is equal to \(1\), therefore so is the right hand side R1
that is \(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}} = – 1\) A1
it follows from the converse to Menelaus’ theorem that \({\text{G, H, I}}\) are collinear R1
[8 marks]