{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (2)
This is another set of tasks where we look at properties derived from the circle theorems. In fact, we continue our look at the properties of the cyclic quadrilateral. Who knew that it could be so interesting!
MONDAY: This is quite a straightforward task, though it might seem odd that the two quadrilaterals don't turn out to be similar.
As corresponding sides of the two quadrilaterals are parallel, their corresponding angles will be equal. So, just as in the smaller quadrilateral, opposite angles of PQRS will be supplementary, which is a sufficient condition for it to be cyclic.
However, as we can plainly see, the two quadrilaterals are not similar, despite their corresponding angles being equal. One might have casually thought that the particular way in which the second quadrilateral is generated from the first, might cause them to be similar, and it is perhaps quite difficult to construct an argument to show that in this situation they don't have to be. On the other hand, it is quite easy to see that, in general, given a cyclic quadrilateral, we can translate one of the sides to construct another quadrilateral (A'B'CD, below) that is plainly not similar but which is still cyclic because the angles are unchanged.
TUESDAY: Here we again have two cyclic quadrilaterals where corresponding angles are equal. Are the quadrilaterals similar this time?
We know that in our two cyclic quadrilaterals, the angles at A and A', and at B and B', are equal, because of the parallel lines. Do students see that this means that the other corresponding angles are equal too? And can they explain why?
It turns out that the two quadrilaterals are not similar, despite having equal corresponding angles. Why? One way to justify this is to consider the segments formed by the chords AB and A'B' (see Task 03A). The segments would need to be similar, but plainly they are not (or if they happened to be similar, we could nullify this by translating one of the chords).
WEDNESDAY: Here we have a trapezium (which takes us back to Week 4) and a quadrilateral with the same corresponding angles as a cyclic quadrilateral but whose shape is rather extreme - can it nonetheless be cyclic?
Quadrilateral FADE is a trapezium. It can't be cyclic as it is not isosceles (only then would its opposite angles be supplementary). We explored this situation in Week 4.
The angles of quadrilateral FBCE are the same as the corresponding angles of the cyclic quadrilateral ABCD, because of the parallel lines, so it too is cyclic (its opposite angles are supplementary). This may come as a surprise, given its extreme shape and it is worth asking students whether they can visualise the circle that goes through its vertices. Where roughly is its centre? It is also interesting to consider what happens if we extend BA produced and CD produced further to the left. What happens when the lines cross so that E and F 'flip over'? Is quadrilateral FBCE (or EBCF?) still cyclic?
THURSDAY: Here we have four circles each containing a cyclic quadrilateral (which might be worth making more visible by drawing the chords JC, ID and HE). What can we say about the angles of these quadrilaterals, and hence about the slopes of these chords?
It turns out that the four cyclic quadrilaterals all have equal corresponding angles, though their shapes are quite different. A good place to start is by comparing the angles of JABC and IJCD. We can show that angle A of JABC is equal to angle C of IJCD (Why?) and similarly that angle B of JABC is equal to angle J of IJCD. Continuing with this kind of argument, we can soon establish that AB is parallel to ID, which in turn means that ID is parallel to GF, and so AB is parallel to GF.
FRIDAY: At first, this task looks rather different, as there are no cyclic quadrilaterals in sight. However, we can think of this as involving two overlapping circles, having cyclic quadrilaterals with a common chord, that have been pulled apart until the circles only touch and the common chord has become a single point - or a tangent.
If we think of the circles as overlapping (below, left) we get the situation in Task 05D, where we have shown that the brown (formerly green) lines are parallel. Alternatively, we can add a tangent to the given diagram (below, right) and use the theorem about the angle between a tangent and chord to show the brown lines are parallel.
The use of 'tangent and chord' is shown below, left, while a nice alternative proof is shown on the right. Both were posted on Twitter.
EXTRA 1: Here is a simple variant on Friday's task. The circles still touch but this time one circle is inside the other. Can we apply the same arguments as before?
EXTRA 2: Finally, a somewhat different task about cyclic quadrilaterals. Here one circle goes through the centre of another, which is a situation that we focus on in next week's tasks.
