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- Apr 14, 2013

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We have the points $Z=(-1,1)$, $A=(-1,6)$ and $B=(3,4)$.

Let $\delta$ be the rotation with center $Z$ and $\delta (A)=B$.

Let $C$ be the point on the circumcircle of the triangle $ABZ$ such that the segment $\overline{CZ}$ goes through the center of circumcircle.

Let $\gamma$ be a rotation with center $C$ and $\gamma (B)=A$.

Show that $\gamma\circ\delta$ is a point reflection at $A$.

To show that do we have to show that there is exactly one fixed point? Or do we have to determine the matrices of the transformation and consider th determinant of the matrix?