{Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} (1)
This is the first of several Weekly sets of tasks looking at properties derived from the circle theorems. In this Week (and the next) we focus on the cyclic quadrilateral, in particular the property that opposite angles of a cyclic quadrilateral are supplementary (ie, their sum is 180˚).
MONDAY: The given quadrilateral is not cyclic as opposite angles do not sum to 180˚. Could we move D (to a point D') in such a way that the angle at D increases from 95˚ to 100˚? If so, this transformed quadrilateral ABCD' would be cyclic: the angles at D' and B would sum to 180˚ (and so of course would the other pair of angles at A and C, as the total sum of the interior angles of a quadrilateral is 360˚).
We can increase the angle at D by moving D 'inwards', towards B. So D lies outside the circle through A, B and C (and D').
TUESDAY: This is very much a visualising task. What sort of size is the circle through A, D and C? Roughly where is its centre? (Note that the given quadrilateral is a trapezium.)
It is not essential to know the location of the circle's centre, though some students will find it helpful: it lies on the perpendicular bisector of the chords AD and DC. The key, perhaps, is to realise that the circle is symmetrical about the perpendicular bisector of DC: point B is closer to this line of symmetry than point A, so B lies inside the circle.
WEDNESDAY: Here we meet the same trapezium again. The task is quite challenging and needs calm and careful thought! You might prefer to go straight to Thursday's task, though that has challenges of its own!
THURSDAY: This is a much more open version of Wednesday's task, involving the same trapezium. Initially, students might try to split the trapezium with a steepish line, ie one that goes through AB and DC. This can be quite intriguing, though it doesn't lead to a solution. It produces quadrilaterals where two pairs of adjacent angles are supplementary, which would allow one of the resulting quadrilaterals to be cyclic, but only if it were an isosceles trapezium. In turn, this would mean that the other quadrilateral had opposite angles of 60˚ and 80˚, and 100˚ and 120˚, but these pairs don’t sum to 180˚ and so this other quadrilateral is not cyclic.
The key is to draw a flatish line, ie one that goes through AD and BC.
Any line such as the green line (below) will do, where angle DPQ = 80˚ and where P is between A and D and Q is between B and C. (What happens if, say, P is between A and D but Q is on CB produced?)
FRIDAY: This interrogates the idea that may have arisen in Thursday's task, of splitting our given trapezium with a steepish line through DC and AB. The two resulting points can't both be the points where the two circles intersect. More specifically, if the two circles intersect at U and V, with U on DC, then V cannot be on AB.
If V were on AB, then the circles can't go through U, C, B and V, and through U, D, A and V, as UCBV and UDAV can't both be cyclic. In the case of UCBV, say, the adjacent angles C and B would be supplementary, as would U and V. If in addition, opposite pairs were supplementary, the shape would have to be an isosceles trapezium, in which case the other quadrilateral would be a (non-rectangular) parallelogram and thus not cyclic.
The two circles will look something like this (below). Point V is not on AB.
How do students solve this? A promising grounded approach is to think of the radius OA (where O is the centre of the circle) as the hand of a clock, and to consider what happens to the angles in, for example, triangle OAD as the hand turns.
Here is an interesting diagram (posted on Twitter by @aburazi1956), where the argument is that the angle at A doesn't change but AO rotates through 6˚, so angle AOA' = 6˚ and so the angles ABA' and ADA', also on the arc AA', will both be 3˚.
