Prove that \(\triangle AGF \,|||\, \triangle ABC\)
About this question
Question 11.3 of Mathematics Grade 12 Paper 2, NSC Examination (DBE, November 2024). It tests the Term 3 ATP topic Euclidean Geometry.
- TermTerm 3
- ATP topicEuclidean Geometry
- PaperPaper 2
- Marks4 marks
\(A\), \(B\), \(G\) and \(F\) lie on the larger circle. A smaller circle touches the larger circle internally at \(A\). \(EA\) is a common tangent to both circles. \(EBCF\) is a tangent to the smaller circle at \(C\). \(AC\) is produced to \(G\). \(AF\) cuts the smaller circle at \(D\). \(AB\), \(CD\) and \(GF\) are drawn.
Diagram labels: at \(A\), \(\hat{A}_1 = \angle EAB\), \(\hat{A}_2 = \angle BAC\), \(\hat{A}_3 = \angle CAD\). At \(C\), \(\hat{C}_1 = \angle ACB\), \(\hat{C}_2 = \angle ACD\), \(\hat{C}_3 = \angle DCF\), \(\hat{C}_4 = \angle FCG\). At \(D\), \(\hat{D}_1 = \angle ADC\), \(\hat{D}_2 = \angle CDF\). At \(F\), \(\hat{F}_1 = \angle AFC\), \(\hat{F}_2 = \angle CFG\). At \(B\), \(\hat{B}_1 = \angle ABE\), \(\hat{B}_2 = \angle ABC\).
In \(\triangle AGF\) and \(\triangle ABC\):
\(A\hat{G}F = \hat{B}_2\) (\(\angle\)s in the same seg)
\(A\hat{F}G = \hat{C}_1 = x\) (proved in 11.1)
\(\hat{A}_3 = \hat{A}_2\) (sum of \(\angle\)s in \(\triangle\))
\(\therefore \triangle AGF \,|||\, \triangle ABC\) (\(\angle\angle\angle\))
More Questions On Euclidean Geometry
In \(\triangle PQR\): \(\dfrac{QE}{QR} = \dfrac{PW}{PR}\) (line \(\parallel\) one side of \(\triangle\); \(PQ \parallel WE\))
\(\therefore PR = \dfrac{PW \cdot QR}{QE}\)
\(P\hat{F}M = 90^\circ\) (tan \(\perp\) diameter)
\(\hat{M} = 180^\circ - 90^\circ - 64^\circ\) (sum of \(\angle\)s of \(\triangle\), in \(\triangle OFM\))
\(\hat{M} = 26^\circ\)
\(OV : OT = 1 : 4\), so \(OV : VT = 1 : 3\)
\(RO \parallel SW\) (from 10.2.1), so in \(\triangle TRO\): \(\dfrac{RS}{ST} = \dfrac{OV}{VT} = \dfrac{1}{3}\) (line \(\parallel\) one side of \(\triangle\))
\(\dfrac{RS}{15} = \dfrac{1}{3}\), so \(RS = 5\)
\(PR = RS = 5\) (line from centre \(\perp\) to chord)
\(PT = 5 + 5 + 15\)
\(PT = 25\) units
- \(\hat{R}_2 = 90^\circ\) (line from centre to midpt of chord)
- \(\hat{K}_3 = 90^\circ\) (given \(AS \perp BD\))
- \(\hat{R}_2 + \hat{K}_3 = 180^\circ\)
- \(\therefore\) OKSR is a cyclic quadrilateral (converse opp \(\angle\)s of cyclic quad)
OKSR is cyclic.
Construction: draw diameter \(CF\) and join \(FB\).
- \(F\hat{C}E = 90^\circ\) (tan \(\perp\) radius)
- So \(B\hat{C}E = 90^\circ - F\hat{C}B\)
- \(F\hat{B}C = 90^\circ\) (\(\angle\) in semi-circle)
- So \(\hat{F} = 90^\circ - F\hat{C}B\) (sum of \(\angle\)s in \(\triangle\))
- \(\therefore B\hat{C}E = \hat{F}\)
- But \(\hat{F} = \hat{A}\) (\(\angle\)s in the same segment)
- \(\therefore B\hat{C}E = \hat{A}\)