Question

1. a rotation by angle ace using point c as the center takes triangle cba onto triangle cde.

a. explain why the image of ray ca lines up with ray ce.
b. explain why the image of a coincides with e.
c. is triangle cba congruent to triangle cde? explain your reasoning.

Explanation:

a. Rotation preserves angle measures. Since it's a rotation about point \(C\), the angle between \(CA\) and \(CE\) is the rotation angle. So, the rotation aligns \(CA\) with \(CE\) as rotation is a rigid transformation that maps one ray to the other.
b. Rotation is a rigid transformation. Point \(A\) is rotated to \(E\) (from part a). The rotation about \(C\) by \(\angle ACE\) moves \(A\) to \(E\) because rotation around \(C\) changes the position of \(A\) such that \(CA = CE\) (rotation preserves distance) and the angle between \(CA\) and \(CE\) is the rotation angle.
c. Rotation is a rigid transformation. Rigid transformations (like rotation) preserve side - lengths and angle measures. So, \(CB = CD\), \(BA=DE\), \(CA = CE\) (from rotation properties: \(r_{\angle ACE,C}(A)=E\), \(r_{\angle ACE,C}(B) = D\), \(r_{\angle ACE,C}(C)=C\)). By SSS (Side - Side - Side) congruence criterion (\(CB = CD\), \(BA = DE\), \(CA=CE\)), \(\triangle CBA\cong\triangle CDE\).

Answer:

a. Rotation is a rigid transformation. It maps \(CA\) to \(CE\) as the rotation about \(C\) by \(\angle ACE\) changes the orientation of \(CA\) to \(CE\) while keeping the length \(CA = CE\).
b. Rotation about \(C\) by \(\angle ACE\) moves \(A\) to \(E\) (as \(CA = CE\) and the angle between them is the rotation angle). Since rotation is a rigid transformation, the image of \(A\) (after rotation) coincides with \(E\).
c. Yes. Because rotation (\(r_{\angle ACE,C}\)) is a rigid transformation. So \(CB = CD\), \(BA = DE\), \(CA=CE\). By SSS, \(\triangle CBA\cong\triangle CDE\).