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 $OO_1 = OO_2 = OO_3$.
The drawing in two steps:
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Description:
On the left diagram, there are three rhombi and the cubish figure is a projection of a $3D$-cube on the plane.
On the right diagram, the additional three dotted lines are the other three edges of the cube invisible to us, yet they also make three more rhombi. Hence the red circle has the same radius.
Consider congruent circles, centres Osub1 Osub2 Osub3 arranged clockwise. O is their common point of intersection. Circle Osub1 and Osub3 intersect at A; Osub 1 and Osub2 intrsect 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 = Osub3C. APCOsub3 is a parallelogram, so AOsub3 = PC. Therefore APCOsub3 is a rhombus. Similarly, AOsub1BP and PBOsub2C are rhombi. Therefore, P is the centre of circle ABC.
