Question

2 here is a diagram.
select all true statements about the figure.
a ( c + b = d + c )
b ( d + b = 180 )
c rotate clockwise by angle ( abc ) using center ( b ). then angle ( cbd ) is the image of angle ( abe ).
d rotate ( 180^{circ} ) using center ( b ). then angle ( cbd ) is the image of angle ( eba ).
e reflect across the angle bisector of angle ( abc ). then angle ( cbd ) is the image of angle ( abe ).
f reflect across line ( ce ). then angle ( cbd ) is the image of angle ( eba ).

Explanation:

Step1: Analyze option A

If \(c + b=d + c\), then subtracting \(c\) from both sides gives \(b = d\). But there is no information in the diagram to suggest \(b=d\). So option A is false.

Step2: Analyze option B

\(d\) and \(b\) are not supplementary. Angles \(d\) and \(a\) are supplementary (\(d + a=180\)) and angles \(a\) and \(b\) are vertical angles (\(a = b\)). So \(d + b
eq180\). Option B is false.

Step3: Analyze option C

When we rotate clock - wise by angle \(ABC\) (which is \(b\) degrees) around center \(B\), angle \(ABE\) (which is \(d\) degrees) is not mapped to angle \(CBD\) (which is \(b\) degrees) since \(d
eq b\). Option C is false.

Step4: Analyze option D

When we rotate \(180^{\circ}\) around center \(B\), angle \(ABE\) is mapped to angle \(CBD\) because a \(180^{\circ}\) rotation around a point swaps the positions of points on a line. Option D is true.

Step5: Analyze option E

If we reflect across the angle bisector of angle \(ABC\), since \(a = b\) (vertical angles) and \(d + a=180\), \(d + b = 180\). The angle bisector of \(ABC\) (angle \(b\)) does not map angle \(ABE\) (angle \(d\)) to angle \(CBD\) (angle \(b\)). Option E is false.

Step6: Analyze option F

When we reflect across line \(CE\), angle \(ABE\) and angle \(CBD\) are congruent (because \(a = b\) (vertical angles) and \(d + a=180\), \(d + b = 180\), and line \(CE\) is a straight line). Reflection across line \(CE\) maps angle \(ABE\) to angle \(CBD\). Option F is true.

Answer:

D. Rotate \(180^{\circ}\) using center \(B\). Then angle \(CBD\) is the image of angle \(EBA\); F. Reflect across line \(CE\). Then angle \(CBD\) is the image of angle \(EBA\).