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 ${\rm{PL}} \times {\rm{PF}} = {\rm{PM}} \times {\rm{PG}}$ .

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]

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.