Question

point d is rotated 180 degrees using b as the center. explain why the image of d must lie on the ray ba.

Explanation:

Step1: Recall 180° rotation property

A 180° rotation about a center \( B \) means that for a point \( D \), its image \( D' \) satisfies \( B \) is the midpoint of \( \overline{DD'} \), and \( \angle DBD' = 180^\circ \). So \( D \), \( B \), and \( D' \) are collinear, with \( B \) between \( D \) and \( D' \).

Step2: Analyze collinearity with ray \( BA \)

From the diagram, \( D \) lies on a line passing through \( B \) and \( A \) (since \( D \) is on the line \( AD \) with \( B \) on it). After a 180° rotation about \( B \), \( D' \) must be on the same line as \( D \) and \( B \) (because 180° rotation preserves collinearity with the center). Also, since the rotation is 180°, the direction from \( B \) to \( D' \) is opposite to \( B \) to \( D \). Since \( D \) is on the side of \( B \) opposite to \( A \) (from the diagram), rotating \( D \) 180° about \( B \) will place \( D' \) on the ray \( BA \) (the ray starting at \( B \) and going through \( A \)), as \( B \) is between \( D \) and \( D' \), and \( D \) is on the line \( BA \) extended beyond \( B \), so \( D' \) will be on \( BA \) (the ray from \( B \) through \( A \)).

Answer:

When a point \( D \) is rotated \( 180^\circ \) about center \( B \), the image \( D' \) satisfies \( B \) is the midpoint of \( \overline{DD'} \) and \( D \), \( B \), \( D' \) are collinear (by the definition of \( 180^\circ \) rotation: a \( 180^\circ \) rotation maps a point to another such that the center is the midpoint, and the three points are colinear with the center between them). From the diagram, \( D \) lies on the line containing \( B \) and \( A \) (line \( AD \)). Since \( D \) is on one side of \( B \), rotating \( D \) \( 180^\circ \) about \( B \) reverses its direction relative to \( B \), so \( D' \) must lie on the ray \( BA \) (the ray starting at \( B \) and passing through \( A \), as \( B \) is between \( D \) and \( D' \), and \( D \) is on the line \( BA \) extended beyond \( B \), so \( D' \) is on \( BA \)).