| 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 \({\rm{HE}} = {\rm{EP}}\) .
The line (AH) meets [BC] at D.
(i) By considering cyclic quadrilaterals show that \({\rm{C}}\widehat {\rm{A}}{\rm{D}} = {\rm{E}}\widehat {\rm{F}}{\rm{H}} = {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) .
(ii) Hence show that [AD] is perpendicular to [BC].
Markscheme
(i) CEFB is cyclic because \({\rm{B}}\widehat {\rm{E}}{\rm{C}} = {\rm{B}}\widehat {\rm{F}}{\rm{C}} = {90^ \circ }\) R1
([BC] is actually the diameter)
(ii) consider the triangles CHE, CPE M1
[CE] is common A1
\({\rm{H}}\widehat {\rm{E}}{\rm{C}} = {\rm{P}}\widehat {\rm{E}}{\rm{C}} = {90^ \circ }\) A1
\({\rm{P}}\widehat {\rm{C}}{\rm{E}} = {\rm{P}}\widehat {\rm{B}}{\rm{A}}\) (subtended by chord [AP]) A1
\({\rm{P}}\widehat {\rm{B}}{\rm{A}} = {\rm{F}}\widehat {\rm{C}}{\rm{E}}\) (subtended by chord [FE]) A1
triangles CHE and CPE are congruent A1
therefore \({\rm{HE}} = {\rm{EP}}\) AG
[7 marks]
(i) EAFH is a cyclic quad because \({\rm{A}}\widehat {\rm{E}}{\rm{B}} = {\rm{C}}\widehat {\rm{F}}{\rm{A}} = {90^ \circ }\) M1
\({\rm{C}}\widehat {\rm{A}}{\rm{D}} = {\rm{E}}\widehat {\rm{F}}{\rm{H}}\) subtended by the chord [HE] R1AG
CEFB is a cyclic quad from part (a) M1
\({\rm{E}}\widehat {\rm{F}}{\rm{H}} = {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) subtended by the chord [EC] R1AG
(ii) \({\rm{A}}\widehat {\rm{D}}{\rm{C}} = {180^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} - {\rm{D}}\widehat {\rm{C}}{\rm{A}}\) M1
\( = {180^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} - (90 - {\rm{E}}\widehat {\rm{B}}{\rm{C)}}\) A1
\( = {90^ \circ } - {\rm{C}}\widehat {\rm{A}}{\rm{D}} + {\rm{E}}\widehat {\rm{B}}{\rm{C}}\) A1
\( = {90^ \circ }\) A1
hence [AD] is perpendicular to [BC] AG
[8 marks]
