Given a circle c(D,r) let {c1(A,r), c2(B,r), c3(C,r)} be three other circles with radius equal to the radius of (c) and centers on (c). Then the second intersection points of these circles {F,G,H} are on a circle of radius r too.
Angle(HCF) is equal to angle(ADB). For this compare the angles with angle(ACB). This implies that HF is parallel and equal to AB. Analogously show the equality of the other sides of triangles ABC and FHG from which the claim follows.
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Four equal circles ![[0_0]](FourEqualCircles_files/FourEqualCircles_0_0_0.png)
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