Another Simple Proof of Johnson's Theorem

Other proofs can be found in cut-the-knot.org. See also Johnson's Three Circles Theorem Revisited and Johnson's theorem proof.

Johnson's theorem: Let three equal circles with centers $J_a$, $J_b$, and $J_c$ intersect in a single point $H$ and intersect pairwise in the points $A$, $B$, and $C$. Then the circumcircle of the triangle $\triangle{ABC}$ is congruent to the original three.

Proof. Since $\odot{ABH}$ and $\odot{BCH}$ are congruent and as $\angle{BAH}$ and $\angle{BCH}$ are angles subtended by the same arc, it follows that $\angle{BAH}=\angle{BCH}$. Analogously, $\angle{CAH}=\angle{CBH}$. Let $O$, $J_a$ be the centers of $\odot{ABC}$ and $\odot{BCH}$, respectively, then

$$\angle{COB}=2\angle{CAB}=2\angle{CAH}+2\angle{BAH}=2\angle{CBH}+2\angle{BCH}=\angle{BJ_aC}.$$

It follows that $\triangle{BCO}$ and $\triangle{BCJ_a}$ are similar isosceles triangles sharing a common side, $BC$, so by $ASA$ we deduce that $\triangle{BCO}\cong{\triangle{BCJ_a}}$. Consequently, $\odot{ABC}$ is congruent to the original three. 

$\square$

Note: The point $H$ may cross the side lines of the triangle $\triangle{ABC}$ in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.