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
Related pages in this website:
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
Triangles -- a home page for anything
about triangles and other polygons.
References
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.