THREE
Some Circle Theorems
1. The Theorem of Ptolemy
Fig. 23
(i) A quadrilateral inequality
THE PROBLEM. ABC is a given triangle and D an arbitrary point. (The diagram shows a typical position of D; other cases are possible, but the differences are slight.) A triangle ABU is constructed as in the diagram so that
We are to prove that
and hence that
THE DISCUSSION.
Also
Hence, in ΔACU, ΔADB,
and
so that
Moreover, from the similar triangles,
and
Hence
(ii) The cyclic case; Ptolemy’s theorem.
The final step of the preceding work,
presupposed that U was not on BC, a condition that would certainly hold in general. The case of exception must now receive attention.
THE PROBLEM. Suppose that, in the preceding work, U ∈ BC. Then
so that
It is required to prove that
[Note the double arrow ⇔.]
Suppose, first, that equality holds, so that U ∈ BC. Then
Hence A, B, C, D are concyclic.
Suppose, next, that A, B, C, D are concyclic. Then
so that BU, BC are the same lines.
Hence U ∈ BC, so that the equality holds.
Note. The four points A, B, C, D can be split in two pairs in three ways,
Ptolemy’s theorem asserts that, when the four points are con-cyclic, the sum of products from two of these pairs is equal to the third. The pair which comes “third” is that defined by the diagonals of the cyclic quadrilateral.
[For a more detailed discussion of implications, see E. A. Maxwell, Fallacies in Mathematics, Cambridge University Press (1959) p. 28.]
Problems
1. If D is a point on the arc opposite A of the circumcircle of an equilateral triangle ABC, then AD = BD + CD.
2. AB, PQ are parallel chords of a circle. Prove that
3. Identify the well-known result which is a special case of the theorem of Ptolemy when ABCD is a rectangle.
4. In ΔABC, draw BE ⊥ AC, CF ⊥ AB. Prove that BF = BC cos B, CF = BC sin B, and write down similar expressions for CE, BE.
By applying the theorem of Ptolemy ...