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start with a 3 4 5 right triangle; ABC.

construct a circle with the diameter along the 3 side. (AB)

it will intersect the side of length 5 at point D.

according the the secant tangent property, we have BC^2 = AD*DC

now construct anther circle with diameter along the 4 side. (BC)

this circle will also intersect the side of length 5 at D.

therefore by the same property, we have AB^2 = AD*DC.

therefore AB^2 = BC^2, or AB = BC or 3 = 4.

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