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Prove any isometry of R^2 is bijection and its inverse is an isometry.

Prove any isometry of R^2 is bijection and its inverse is an isometry.

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Answer #1

Let ABC be the triangle with vertices (1,0), (0,1), and (0,0). Let f be an isometry as defined in the post. Suppose f takes A, B, and C to A?, B?, and C?.

There is a combination ? of rotation and/or reflection and/or translation that takes ABC to A?B?C?. Then ??1?f is an isometry as defined in the post. Note that it leaves A, B, and C fixed.

Given an unknown point P=(x,y), if we know the distances from P to A, B, and C, we know x and y. Since ??1?f fixes A, B, and C, it is the identity. Thus f=?.

Rotations, reflections, and translations are surjective.

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