It Stands to Reason
Geometry and Depth
How do we provide opportunities for our pupils to develop both fluent technical skills and deep conceptual understanding and also the connections between these (as is demanded by the new National Curriculum for Mathematics), alongside developing their mathematical reasoning and problem solving confidence and resilience? This month we consider one possible activity related to the geometry of triangles that provides such an opportunity to go deeper, developing the skills and habits described explicitly in the National Curriculum: the activity entails pupils following a line of enquiry, conjecturing relationships and generalisations, and then developing an argument, justification or proof using mathematical language which they apply to this nonroutine problem. To do so, they will break down the problem into a series of simpler steps, and will (hopefully!) persevere in seeking a solution.
So what is the activity? It’s a classic of triangle geometry: demonstrate that the three medians of a triangle intersect at a common point, where they divide each other in the ratio 2:1.
Part 1 – Illustrate the problem
All geometrical problems need a visualisation. This could be done by hand or with the assistance of graphing/geometry software. A good picture gives pupils the opportunity to develop a visual representation of the problem and also the notation needed to help both description and later reasoning (Fig 1).
Fig 1 (click to enlarge)
The triangle has been labelled formally ABC as have the midpoints D, E and F of each edge, and each corner has been joined to the opposite midpoint to make the medians. Here, of course, we have not assumed that the three medians are in fact concurrent at G.
Part 2 – Simplify the problem
To deduce the result using their reasoning skills from already known and proven results (like similar triangles or properties of 2D shapes) your pupils probably need to simplify the set up first. This could be attempted in a number of different ways, usually simplifying the problem to initially considering only two of the medians at once (Fig 2).
Fig 2 (click to enlarge)
Part 3 – Solve the problem
Pupils should already know about and be confident with reasoning with similar triangles before going deeper with a problem proof like this. Once they do, this will probably be a rich challenge for them. They may need some scaffolding/structure (like this) to get started:
Stage 1: Show that triangle CAB and triangle CED are similar triangles (and identify the correspondence and the scale factor).
Stage 2: Deduce that AB is parallel to ED and that AB = 2ED.
Stage 3: Show that triangle AGB is similar to triangle DGE (and identify the correspondence and the scale factor).
Stage 4: Deduce that AG = 2GD and BG = 2GE.
Stage 5: Interpret Stage 4 in relation to what we are trying to prove.
By this stage, they have established that G is the point of intersection of two of the medians and that it divides them in the ratio 2:1, but they will need to do more to justify that all three medians are concurrent at G. So, they pick another pair of medians (Fig 3). This is the bit that develops resilience!
Fig 3 (click to enlarge)
Notice that, again, they must not assume that the new medians intersect at G; we have given the new intersection the name H. Scaffolding should now be less likely to be required, if it was provided at the beginning.
Stage 6: Repeat Stages 1 to 5 for the new pair of medians, being careful to adapt notation accordingly.
Stage 7: Interpret Stage 6 in relation to what we are trying to prove.
Stage 8: Deduce that the three medians all intersect at a single point G, and that G divides each median in the ratio 2:1 as required.
Stage 9: Smiles and “phew!” all round.
Throughout, an emphasis will need to be placed on what your pupils record, how they record it, and the order in which they record it. One idea is to give them the appropriate formal steps but not in the right order for Stages 1 to 5 and ask them to put them in the correct logical order, and then get them to replicate this argument themselves in Steps 6 to 8.
Part 4 – An Alternative Approach
As an alternative, some pupils may prefer/come up with an alternative approach using both similar triangles and properties of other shapes (Fig 4).
Fig 4 (click to enlarge)
Here the scaffolding might look like this, with the option to dispense with some or all of it and/or to provide the formal written steps for some or all stages but out of logical order for your pupils to reorder correctly.
Stage 1: Show that triangle CAB and triangle CED are similar triangles (and identify the correspondence and the scale factor) – as before.
Stage 2: Deduce that AB is parallel to ED and that AB = 2ED – as before.
Stage 3: Mark in L and M, the midpoints of AG and BG. Deduce that DELM is a parallelogram (opposite sides are equal and parallel) and so G bisects its diagonals (ideally, this “subtheorem” would be proved too).
Stage 4: Further deduce that G divides AD and BE in the ratio 2:1.
Stage 5: Interpret Stage 4 in relation to what we are trying to prove.
Stage 6: Repeat Stages 1 to 5 (resilience!) for a new pair of medians being careful to adapt notation accordingly.
Stage 7: Interpret Stage 6 in relation to what we are trying to prove.
Stage 8: Deduce that the three medians all intersect at a single point G, and that G divides each median in the ratio 2:1 as required.
You might at this point want to “tick off” the curriculum points that this activity addresses. I’d suggest that it covers at least:
Developing fluency
 use language and properties precisely to analyse 2D shapes
Reasoning mathematically
 begin to reason deductively in geometry, including using geometrical constructions
 extend and formalise knowledge of ratio and proportion in working with geometry, and in formulating proportional relations
Solving problems
 develop mathematical knowledge including multistep problems
 develop use of formal mathematical knowledge to interpret and solve problems
 begin to model situations mathematically and express the results using a range of formal mathematical representations
 select appropriate concepts, methods and techniques to apply to unfamiliar and nonroutine problems
Ratio, proportion and rates of change
 express the division of a quantity into two parts as a ratio
 understand that a multiplicative relationship between two quantities can be expressed as a ratio or a fraction
 relate the language of ratios and the associated calculations to the arithmetic of fractions
Geometry and measures
 describe, sketch and draw using conventional terms and notations: points, lines, parallel lines
 use the standard conventions for labelling the sides and angles of triangle ABC, and know and use the criteria for congruence and similarity of triangles
 derive and illustrate properties of triangles [for example, equal lengths and angles] using appropriate language and technologies
 apply the properties of angles at a point, angles at a point on a straight line, vertically opposite angles
 derive and apply formulae to calculate and solve problems involving area of triangles
 understand and use the relationship between parallel lines and alternate and corresponding angles
 apply angle facts, triangle congruence, similarity and properties of quadrilaterals to derive results about angles and sides and use known results to obtain simple proofs.
Not bad for one activity!
Part 5  A Possible Extension
What else could you do with this triangle and its medians now that you have established concurrence and position of the intersection? One suggestion is to start to look at the areas of the triangles formed by the medians (Fig 5) and ask your pupils to establish relationships between the areas of the various triangles using the standard results that: 1) triangles on the same base with the same height have the same area, and 2) triangles on different bases with the same height have areas in the ratio of the bases. Both of these, especially the second, are worthwhile for your pupils to prove; again, they may need some scaffolding.
Fig 5 (click to enlarge)
The goal is that the triangles 1 to 6 are fully justified as having equal areas. This then leads to investigating relationships between the areas of the any of the six small triangles and the areas of any of the larger triangles, including the original whole triangle ABC (which is six times larger than any of triangles 16).
Part 6  More Advanced Links
Ceva’s Theorem (not generally taught in most schools or colleges these days, regrettably) provides a more general result about concurrent lines and can itself be justified using a more technical ‘if and only if’ style proof. It can be applied to demonstrate concurrence of the medians of a triangle very quickly, and also to prove similar results for the concurrence of the altitudes and angle bisectors of any triangle.
Those who teach Centres of Mass at KS5 (and find that their students have all but forgotten their basic geometry and geometrical reasoning by Y12/13 in the midst of the algebraic geometry approach now prevalent in AS/A Level Mathematics and Further Mathematics) will also find this activity a useful reminder for students when justifying the location of the centroid/centre of mass/centre of gravity of a uniform triangular lamina.
However or whenever you use this activity, it will certainly provide a rich opportunity for your students to further and hone their geometrical reasoning skills.
Image credit
Page header by Japanexpertna.se (adapted), some rights reserved
