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?
 A: This is based on the Inscribed Angle Theorem, which says that an inscribed angle measures half the angle of the arc that it subtends on the circle. The following proof is fairly simple.
Let $D$ be the center of the circle. Note that $\angle DAC=\angle DCA$ and that 
$$
\angle DAC+\angle DCA+\angle ADC=\pi=\angle ADC+\angle CDE\tag{1}
$$
$\hspace{3.2cm}$
Subtracting $\angle ADC$ from both sides of $(1)$ and remembering that $\angle DAC=\angle DCA$, we have
$$
\angle CDE=2\angle DAC
$$
Similarly,
$$
\angle BDE=2 \angle DAB
$$
Taking a sum or difference, we get that
$$
\angle CDB=2\angle CAB
$$
$\square$
In your diagram, this means that $\angle ACB$ is half the arc from $A$ to $B$. The same is true of $\angle ADB$.

To subtend the same arc of a circle, the vertices of the angles must be on the same side of the chord between the ends of the arc. To prove the converse, we need to assume that the points considered be on the same side of the chord.
Suppose we have a point $C$ on a circle with a given $\angle ACB$. Suppose that $\angle ADB$ is the same angle, but $D$ is not on the circle.
$\hspace{3.2cm}$
Find the point $E$ at the intersection of $\overline{BD}$ and the circle. We know by the Inscribed Angle Theorem, that $\angle AEB=\angle ACB$. However, if $D$ is outside the circle, $\angle AEB\gt\angle ADB$, and if $D$ is inside the circle, $\angle AEB\lt\angle ADB$. Therefore, $D$ must be on the circle.
$\square$
Thus, if two points are on the same side of the chord, and the chord subtends the same angle at both, then they are on the same circle with that chord.
A: 
Think that angles $ABO$ and $DCO$ are equal. Also we know angles $COD$ and $BOA$ are equal. Here we have two results:


*

*Angles $OAB$ and $ODC$ are equal.

*Triangle $BOA$ and $OCD$ are similar.


From the similarity we can know $OB/OC=OA/OD$, also the angles $BOC$ and $AOD$ are equal. Then we can know triangles $OBC$ and $OAD$ are similar. From the similarity and question information we can know angles $OBC=OAD$ $OBA=OCD$  $ODA=OCB$ $ODC=OAB$. The sum of these all equations gives us: $A+C=B+D$.
A: 
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
