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This theorem was proved by Giovanni Ceva (1648-1734).
On the diagonal BD locate a point M such that angles BCA and MCD are equal. Since angles BAC and MDC subtend the same arc, they are equal. (why?) Therefore, triangles ABC and DMC are similar. Thus we get CD/MD = AC/AB, or AB�CD = AC�MD. |
Since angles BCA and MCD are equal, then angle BCM=BCA+ACM equals angle ACD=ACM+MCD. So triangles BCM and ACD are similar which leads to
BC/BM = AC/AD, or BC�AD = AC�BM.
Summing up the two identities we obtain
AB�CD + BC�AD = AC�MD + AC�BM = AC�BD
http://cut-the-knot.com/proofs/ptolemy.html gives this proof along with the interesting fact about the three chords formed by the vertices of an equilateral triangle and any fourth point on the circle: the sum of the shorter two equals the longer.
Summary of geometrical theorems
Inscribed Angles -- a proof that all angles inscribed in a circle subtend an arc that is twice the arc subtended by the same central angle.
Ptolemy's Inequality -- a generalization of Ptolemy's Theorem for convex quadrilaterals that aren't necessarily cyclic
Butterfly Theorem: given a chord PQ with midpoint M, and two other chords AB and CD that intersect at M, then M is the midpoint of the intersections with PQ by AD and CB. Additionally (or alternatively), if AB and CD are considered the diagonals of cyclic quadrilateral ACBD, then opposite sides of ACBD intersect PQ at points equidistant from M.
Triangles -- a home page for anything about triangles and other polygons.
The webmaster and author of this Math Help site is Graeme McRae.
Ptolemy's theorem states that given a cyclic quadrilateral (i.e. one that
can be inscribed in a circle) the product of the diagonals equals the sum of
the products of opposite sides.