Property of cyclic quadriterals proof! http://en.wikipedia.org/wiki/Cyclic_quadrilateral
This article states that: "Another necessary and sufficient condition for a convex quadrilateral ABCD to be cyclic is that an angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.[3] That is, for example, 
$$\angle ACB = \angle ADB$$"
I haven't found a proof for this online, can somebody proof this or find one?
Also, does this mean that if you draw in the diagonals, that of the 4 triangles you create the opposite ones are similar?


The proof is based on the following theorem:

The angles subtended by a chord at the circumference of a circle are
  equal, if the angles are on the same side of the chord

Hence $\angle{ACB} = \angle{ADB}-----------> 1$
similarly $ \angle{CBA} = \angle{CDA} ------------------>2$
Adding 1 and 2 .
$\angle{ACB} + \angle{CBA} = \angle{CDB}$
$ 180^\circ - \angle{CAB} = \angle{CDB}$
$\angle{CAB} + \angle{CDB} = 180^\circ$
This means the quadrilateral is inside a circle because as we know if the opposite angles of a quadrilateral sum to $180^\circ$ the quadrilateral is cyclic(this is a well known proof).
Hence this is a necessary and sufficient condition for the quadrilateral to be cyclic
