As I develop some of the higher level geometry waypoints for our framework I’m widening my understanding of ‘different geometries’ and am trying to offer a balance of approaches and show how they blend together.
We have many ways to describe the Euclidean plane and objects in the plane. Here’s the case of the isosceles triangle, which shows how our starting definitions can be different yet the same properties deduced (admittedly in different ways).
Now I wouldn’t recommend you start looking at this particular case with students as it is here as a base example for you to consider and compare. But what might be useful is the narrative that goes on to explain the properties of the isosceles triangle: there isn’t one unique explanation, but justifications can take on different ‘lenses’, which is worth thinking about when we listen to students’ responses that aren’t quite the way we’d explain it!
Euclid defined an isosceles triangle as a triangle with two equal sides:
What can we deduce from this using the congruence axioms?
If triangle ABC has |AB| = |AC| then triangles ABC and ACB are congruent by side-angle-side (SAS).
In the two triangles above both left hand sides are equal, both right hand sides are equal and the angle at A is the same angle (since they are the same triangle).
It follows from SAS that if two triangles are congruent then all corresponding angles are congruent. In other words the angle at B is equal to the angle at C, so angles opposite the equal sides are equal.
Now bisect the angle at A and draw in line segment AM.
Again by SAS triangles ABM and ACM are congruent; |AB| = |AC|, the line segment AM bisected the angle at A so ∠BAM and ∠CAM are equal, the triangle shared side AM.
Therefore |BM| = |MC|, M is the midpoint of BC and AM is perpendicular to BC. Hence AM is the altitude and median of triangle ABC.
Now another approach is using transformations.
An isosceles triangle is a triangle with at least one line of symmetry. This means that triangle ABC is its own image under reflection in AM.
Logic and my knowledge tells me that this must look like this:
This is written as r(△ABC) = △ACB.
So B is reflected to C, r(B) = C, C to B, r(C) = B and A is on the mirror line, r(A) = A
What other properties can we deduce from this?
a) Since r(B) = C and r(A) = A:
• |AB| = |AC|, two sides are of equal length
b) Since r(B) = C, AM is the mirror line, M lies on BC and r(M) = M:
• M is the midpoint of BC
• AM is the perpendicular bisector of BC
• AM is the median and altitude of the triangle ABC
c) Since r(A) = A, r(B) = C, and r(M) = M:
• r(∠CAM) = ∠CBM • ∠CAM = ∠CBM, the two ‘base angles’ are equal
It’s worth comparing these two arguments. How are they similar or different? Do you have a preference? Can you apply these kinds of arguments to explain the properties of a parallelogram?
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SOMETHING TO TRY
KS1: Where in the classroom do you sit? How many ways can you describe your position?
KS2: In how many different ways can you describe a square? Think about its vertices, edges, and symmetry.
KS3: How could you find the mid-point of a line segment?
In what ways might your method change if:
· The line segment was drawn on a piece of tracing paper?
· The line segment was drawn on squared paper?
· You had a ruler?
· You had the co-ordinates of the ends of the line segment?
· You had the vector describing the journey from one end of the line segment to the other?
KS4: Here are two definitions of a parallelogram. Show how each leads to some properties of a parallelogram using congruency or transformations.
a) a quadrilateral with parallel opposite sides
HINT: first show angle information using parallel angle rules (extend the sides of the parallelogram), then draw in a diagonal, work on the angles and shared side of the two triangles to show ASA congruency to lead to opposite sides being equal in length
b) a quadrilateral with rotational symmetry order 2
HINT: consider the images of vertices and sides of the parallelogram when it is rotated 180˚, consider special properties of any line rotated about any point 180˚
KS5: Investigate the many proofs of Pythagoras’ theorem including each of the following:
Using similar triangles and scale factors of length and hence area:
Using a geometrical diagram and algebraic manipulation, such as investigating the different algebraic representations of the area of this trapezium that is constructed using 3 right angled triangles, two of which are congruent:
Using the properties of shearing. Read through the explanation here or search online for an animation.
For further ideas visit https://www.cut-the-knot.org/pythagoras/ |
There’s no shame in not knowing the words… what is a shame is that we are not always confident to stop and ask for their meaning. Wouldn’t it be great if we could all feel comfortable enough to ask for an explanation; after all it’s what we wish the students would do!
Here are a few from this blog that may or may not be of use.
Midpoint: The point in the centre of a line segment or half way between two points.
Altitude of a triangle: the line segment from a triangles vertex to the opposite side, such that the two line segments are perpendicular. Every triangle has three altitudes.
Median of a triangle: the line segment from a triangles vertex to the midpoint of the opposite side. Every triangle has three medians.
