Johnson's theorem proof as it is in the title. I am have problems with proving the Johnsons Theorem. 

The circles $O_1,O_2,O_3$ are all congruent. I am supposed to show that the circle passing though point A,B,C is also congruent. I am supposed to use the fact that i can draw parallelograms here, rhombi in particular.I see four rhombi but i cannot see where to go from here. I also only know the basic facts about parallelograms so please no intricate theorems.
One proof. A direct, brute force angle chasing is one way to do the job.
Observe that the line through point $O_1$ orthogonal to $AB$ is the orthogonal bisector of $AB$ (because $AB$ is a chord of the circle with center $O_1$). Analogously, the perpendiculars from points $O_2$ and $O_3$ to the edges $CA$ and $BC$ respectively are orthogonal bisectors of $CA$ and $BC$ respectively. These three orthogonal bisectors intersect at a common point $Q$ which is the center of the circumcircle of triangle $ABC$. Let $\angle \, ACO = \alpha$ and $\angle \, OCB = \beta$. Then in the circle with center $O_3$
$$\angle \, AO_3O = 2 \, \angle \, ACO = 2\alpha$$
Quad $OO_3AO_1$ is a rhombus (or if you prefer $AO_3O$ and $AO_1O$ are congruent triangles) so $\angle \, AO_1O = \angle \, AO_3O = 2 \alpha$. 
In the circle with center $O_2$ $$\angle \, OO_2B = 2 \, \angle \, OCB = 2\beta$$
Quad $OO_2BO_1$ is a rhombus (or if you prefer $BO_2O$ and $BO_1O$ are congruent triangles) so $\angle \, OO_1B = \angle \, OO_2B = 2 \beta$. Consequently, $$\angle \, AO_1B = \angle \, AO_1O + \angle \, OO_1B = 2 (\alpha + \beta)$$ Therefore $$\angle \, AO_1Q = \frac{1}{2} \, \angle \, AO_1B = \alpha + \beta$$ However, in triangle $ABC$ point $Q$ is a circmcenter, so $\angle \, AQO_1 = \frac{1}{2} \, \angle \, AQB = \angle \, ACB = \alpha + \beta$. Thus we conclude that 
$$\angle \, AQO_1 = \alpha + \beta = \angle \, AO_1Q$$  which yields that triangle $AQO_1$ is isosceles and $AQ = AO_1$. Since $Q$ is the circumcenter of $ABC$, the segment $AQ$ is the radius of the circumcircle and is therefore equal to the radius $AO_1$ of the circle with center $O_1$ which is what you want to prove.

Comment: On a side note, one can easily show that $O$ is the orthocenter of triangle $ABC$, i.e. $AO$ is orthogonal to $BC$, $\, BO$ is orthogonal to $CA$ and $CO$ is orthogonal to $AB$. 

Another proof. Another way to prove the theorem is by proving that triangles $ABC$ and $O_1O_2O_3$ are congruent (in fact central symmetric to each other) and since $O$ is the circumcenter of $O_1O_2O_3$  the radius of its circumcircle $OO_1 = OO_2=OO_3$ must be equal to the radius of the circumcircle of $ABC$.

The congruence can be proven as follows: Look at quad $BCO_2O_3$. Segment $BO_3$ is equal and parallel to segment $OO_1$ because $BO_1OO_3$ is a rhombus. Similarly, $CO_2$ is equal and parallel to $OO_1$ because $CO_1OO_2$ is a rhombus. Therefore $BO_3$ is equal and parallel to $CO_2$ so $BCO_2O_3$ is a parallelogram and $BC$ is equal an parallel to $O_3O_2$  (and by the way $O_3O_2$ is orthogonal to $AO$ because $AO_3OO_2$ is a rhombus so $BC$ is orthogonal to $AO$). Analogously, one shows that $CA$ is equal an parallel to $O_1O_3$, and $AB$ is equal an parallel to $O_2O_1$. Therefore triangles $ABC$ and $O_1O_2O_3$ are congruent and the radii of their circumcircles are equal. The curcumradius of $O_1O_2O_3$ is $OO_1 = OO_2 = OO_3$.
Johnson's Theorem Diagram

Consider congruent circles, centres $O_1$ $O_2$ $O_3$ arranged clockwise. O is their common point of intersection.
Circle $O_1$ and $O_3$ intersect at A; $O_1$ and $O_2$ intersect at B, and C is third intersection.

The nine radii form three dotted rhombi (with O as the common point). 
Each of the three different sets of three parallel lines has matching dots.
Draw AP // and = $O_3$C. 
APC$O_3$ is a parallelogram, so A$O_3$ = PC. 
Therefore APC$O_3$ is a rhombus.
Similarly, A$O_1$BP and PB$O_2$C are rhombi.
Therefore, P is the centre of circle ABC.