| Date | May 2014 | Marks available | 14 | Reference code | 14M.2.hl.TZ0.9 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
The diagram above shows a point O inside a triangle ABC. The lines (AO), (BO), (CO) meet the lines (BC), (CA), (AB) at the points D, E, F respectively. The lines (EF), (BC) meet at the point G.
(a) Show that, with the usual convention for the signs of lengths in a triangle, BDDC=−BGGC.
(b) The lines (FD), (CA) meet at the point H and the lines (DE), (AB) meet at the point I. Show that the points G, H, I are collinear.
Markscheme
(a) applying Ceva’s theorem to triangle ABC,
CDDB×AEEC×BFFA=1 M1A1
applying Menelaus’ theorem to triangle ABC with transversal (GFE),
BGGC×CEEA×AFFB−=1 M1A1
multiplying the two equations, M1
CDDB×BGGC=−1 A1
so that BDDC=−BGGC AG
[6 marks]
(b) similarly
CEEA=−CHHA M1A1
and AFFB=−AIIB A1
multiplying the three results,
BDDC×CEEA×AFFB=−BGGC×CHHA×AIIB 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 BGGC×CHHA×AIIB=−1 A1
it follows from the converse to Menelaus’ theorem that G, H, I are collinear R1
[8 marks]
