| Date | None Specimen | Marks available | 7 | Reference code | SPNone.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
In the acute angled triangle ABC, the points E, F lie on [AC], [AB] respectively such that [BE] is perpendicular to [AC] and [CF] is perpendicular to [AB]. The lines (BE) and (CF) meet at H. The line (BE) meets the circumcircle of the triangle ABC at P. This is shown in the following diagram.
(i) Show that CEFB is a cyclic quadrilateral.
(ii) Show that HE=EP .
The line (AH) meets [BC] at D.
(i) By considering cyclic quadrilaterals show that CˆAD=EˆFH=EˆBC .
(ii) Hence show that [AD] is perpendicular to [BC].
Markscheme
(i) CEFB is cyclic because BˆEC=BˆFC=90∘ R1
([BC] is actually the diameter)
(ii) consider the triangles CHE, CPE M1
[CE] is common A1
HˆEC=PˆEC=90∘ A1
PˆCE=PˆBA (subtended by chord [AP]) A1
PˆBA=FˆCE (subtended by chord [FE]) A1
triangles CHE and CPE are congruent A1
therefore HE=EP AG
[7 marks]
(i) EAFH is a cyclic quad because AˆEB=CˆFA=90∘ M1
CˆAD=EˆFH subtended by the chord [HE] R1AG
CEFB is a cyclic quad from part (a) M1
EˆFH=EˆBC subtended by the chord [EC] R1AG
(ii) AˆDC=180∘−CˆAD−DˆCA M1
=180∘−CˆAD−(90−EˆBC) A1
=90∘−CˆAD+EˆBC A1
=90∘ A1
hence [AD] is perpendicular to [BC] AG
[8 marks]
