Thankyou

sfwc
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Since you have not changed your argument, it's I who should be distressed for being so obtuse.

Thank you.

Carry out the construction as given to get line PM, but extend this line meet the opposite side of the circle at a new point Q. Extend QB and AC to meet at F. Now triangles MQF and MQA are mirror congruent, since both are right triangles sharing a side and their angles at Q subtend equal arcs (since P bisects the arc AB). Therefore AM = MF and angle QAM = angle AFM.

Angles QAM + MBQ = pi, since they subtend complementary arcs, and MBQ + FBC = pi since they are supplementary angles. Therefore AFM = QBM = FBC, so triangle FBC is isoceles, and so CF = BC. QED

Actually, that last part of the proof is a pretty little fact in its own right: For any inscribed quadrilateral, extending two opposite sides to meet completes it to an isoceles triangle, and the small added triangle is similar to the large one.

--Stuart Anderson

>For any inscribed quadrilateral,
>extending two opposite sides to meet
>completes it to an isoceles triangle,

This is not true, as you can take two points, say A and Q, on a circle that will serve as the base vertices of a triangle with an arbitrary apex F outside the circle so that AF meets the circle in C and QF meets it in B. Just move F so AQBC is not isosceles.

>and the small added triangle is similar
>to the large one.

This is true and is a characterization of cyclic quadrilaterals. In particular, if the property holds for one pair of opposite sides, it holds for the other pair as well.

Yes, I see I was too hasty. Since this was just a closing remark I had added at the last minute, I did not take sufficient thought about my assumptions.

Here is another thought:

In the context of this proof, I started thinking about intersecting chords in circles, and went looking for information about them on CTK to see what properties you had already discussed. There is a general property which I could not find anywhere, although I am sure it is well known, so I will provide a short discussion of it here.

Let chords AB and CD cross at P (inside or outside the circle), and let them meet at a fixed angle theta. Then the sum of the (signed) arcs AC and BD is independent of the location of P.

Proof: Let A'B' be a new chord parallel to AB, meeting CD at P'. Then the arcs AA' and BB' are equal by symmetry around the common perpendicular bisector of AB and B'B'. But one of these arcs is added and one is removed from the arcs AC and BD to form arcs A'C and B'D. Therefore the sum of the arcs remains constant.

Similarly, one can move chord CD to a new position, causing P to slide along AB. The requirement that AB and CD be chords means that P must lie within a circumscribed parallelogram with sides parallel to AB and CD. Within this region, the two chord-sliding operations clearly suffice to move P from any point to any other point, therefore the sum of signed arcs is constant as required. QED

I specifically wanted to prove this without using the theorem that peripheral angles subtending the same arc are equal, because this is a simple consequence of the general result, and I wanted a "pure" proof. On the other hand, this would make a good example of a special case being equivalent to the general case.

General --> Special: Move P first to the center and second to the periphery of the circle. At the periphery, one arc vanishes, say, AC. Then arc BD is equal to the sum of the arcs found with P at the circle center, which are equal by symmetry. Therefore a given angle on the periphery subtents twice the arc of the same angle at the center.

Special --> General: Construct the extra line AD. Then angle BAD subtends arc BD and angle CDA subtends arc AC. Considering the triangle ADP and the two pairs of vertical angles at P shows that angle BAD + CDA = BPD = APC. Therefore by the special case, an angle equal to APC placed at the circle center would subtend an arc equal to half the sum of arcs AC and BD. Thus placing P at the circle center would cause two equal arcs to be subtended, each equal to half AC + BD. Therefore, the arcs subtended by crossed chords at any location P sum to the same value as when P is at the circle center, and hence the sum of arcs AC and BD is constant.

I think this would make a good java animation. You could drag P around and watch the lengths of the subtended arcs. Also, this property seems to invite side-by-side discussion with the similar property about lengths of segments of crossed chords which you have already shown under the heading "crossed chords."

If this has already been discussed on CTK, my apologies for the long post!

--Stuart Anderson

Strange, I could not find this either.

>Let chords AB and CD cross at P (inside or outside the
>circle), and let them meet at a fixed angle theta. Then the
>sum of the (signed) arcs AC and BD is independent of the
>location of P.

Yours is a very nice proof. Thank you.

https://www.cut-the-knot.org/Curriculum/Geometry/SecantAngle.shtml

I called the incident "strange" before because of a clear recollection that something has been written on the topic. I have updated the page and added another with your derivation:

https://www.cut-the-knot.org/Curriculum/Geometry/AnglesInCircle.shtml

Thank you,
Alex

Thank you for making the nice Java demonstration. It looks really good and playing with it makes the proof absolutely obvious, I think. There are a couple of minor points about the jave demo, though:

First, when the indicated angle is > 90 degrees, and the intersection point P is moved very near the edge of the circle, it'subtends an arc > 180 degrees, which causes the animation to change over to coloring the minor arc red instead of continuing to color the major arc blue. This is not quite in line with what you want, I think.

Second, and this is really an oversight on my part, not a problem with the java animation, when the point A is moved outside the circle at a location where only the rays AE and AB meet the circle, but AC and AD do not, then the arcs CB and DE overlap. The theorem remains true of course, but visually it is less obvious.

Thank you,
Stuart

I made some improvements. It certainly better now. Although, at times, depending on the orientation of the lines, there is a glitch I think when A is in a vicinity of the circle. Takes to much time to detect the condition.

Alex