Specialist Mathematics Unit 1
Sample context for assessment – Ceva’s theorem
Introduction
The theorem, was developed by Giovanni Ceva in the late 17th century CE. It relates ratios of
lengths associated with line segments from the vertices of a triangle to the opposite side and their
common point of intersection. An earlier proof was given in the work of the Arab mathematician
Yusuf ibn Ahmad al-Mu'taman ibn Hūd in Spain in the latter part of the 11th century CE. It is also
related to Menelaus' theorem.
Ceva’s Theorem
If ABC is a triangle and line segments are drawn from the vertices to a common point O and then
extended to the opposite side then
BD CE AF
DC  EA  FB = 1.
Where the segments are directed, that is we write for example AB = – BA
Part 1
a.
Use technology to produce various triangles with this construction and check that the theorem
holds for these triangles (for example: http://www.geogebra.org/m/23785 or
http://www.geogebra.org/material/simple/id/127898).
b.
By considering areas of suitable triangles, develop a proof of the theorem.
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graph theory proofs
Part 2
An alternative form of the theorem is given by:
.
sin (ABE)  sin (BCF)  sin (CAD) = sin (CBE)  sin(ACF)  sin (BAD)
Use trigonometry to prove this alternative form.
Part 3
a.
State the converse of the theorem and develop a proof of the converse
b.
Use the converse of the theorem to prove the medians of a triangle are concurrent.
Areas of study
The following content from the areas of study is addressed through this learning activity.
Area of study
Topic(s)
Content dot point
Geometry,
measurement and
trigonometry
Geometry in the plane and proof
1, 2, 3, 4, 5, 6.
Outcomes
The following outcomes, key knowledge and key skills are addressed through this task.
Outcome
Key knowledge dot point
Key skill dot point
1
1, 2, 3, 4
1, 2, 3, 4
2
2, 3, 4
2, 3, 4
3
1, 3, 5
1, 2, 3, 4, 5, 6, 8, 9
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