Question

triangle abc is a reflection of triangle abc across line bc. the statements below are scrambled. rearrange them into a logical proof that bc is the angle bisector of ∠aba. rays ba and ba are mirror images across bc, with b as the vertex. abc is the image of abc under a reflection across line bc. thus m∠abc = m∠cba. a reflection is a rigid motion that preserves angle measure. therefore ∠abc ≅ ∠abc. points b and c lie on the line of reflection so b and c are fixed; a maps to a. if the two angles adjacent to bc at b are equal then bc bisects ∠aba.

Explanation:

Step1: State the reflection

$A'B'C'$ is the image of $ABC$ under a reflection across line $BC$.

Step2: Mention fixed - points

Points $B$ and $C$ lie on the line of reflection, so $B$ and $C$ are fixed; $A$ maps to $A'$.

Step3: Recall property of reflection

A reflection is a rigid motion that preserves angle measure.

Step4: Establish angle - equality

Therefore $\angle ABC\cong\angle A'B'C'$.

Step5: Write angle - measure equality

Thus $m\angle ABC = m\angle CBA'$.

Step6: Define angle - bisector condition

If the two angles adjacent to $BC$ at $B$ are equal then $BC$ bisects $\angle ABA'$.

Step7: State the mirror - image property

Rays $BA$ and $BA'$ are mirror images across $BC$, with $B$ as the vertex.

Answer:

1. $A'B'C'$ is the image of $ABC$ under a reflection across line $BC$.
2. Points $B$ and $C$ lie on the line of reflection, so $B$ and $C$ are fixed; $A$ maps to $A'$.
3. A reflection is a rigid motion that preserves angle measure.
4. Therefore $\angle ABC\cong\angle A'B'C'$.
5. Thus $m\angle ABC = m\angle CBA'$.
6. If the two angles adjacent to $BC$ at $B$ are equal then $BC$ bisects $\angle ABA'$.
7. Rays $BA$ and $BA'$ are mirror images across $BC$, with $B$ as the vertex.