Question

the image of $\triangle abc$ after a reflection across $\overleftrightarrow{eg}$ is $\triangle abc$.
which triangle must be a right triangle and why?
$\bigcirc$ $\triangle abc$ is right because it is the image of $\triangle abc$.
$\bigcirc$ $\triangle adc$ is right because $\overline{aa}$ intersects $\overline{ac}$ at $a$.
$\bigcirc$ $\triangle bcc$ is right because $b$ lies of the line of reflection.
$\bigcirc$ $\triangle bgc$ is right because $\overleftrightarrow{eg} \perp \overline{cc}$.

Explanation:

To determine which triangle is a right triangle, we analyze each option:

• Option 1: A reflection preserves the shape (congruence), but it doesn't guarantee the original triangle was right, so this is incorrect.
• Option 2: The intersection of \( \overline{AA'} \) and \( \overline{AC} \) at \( A \) doesn't imply a right angle, so this is incorrect.
• Option 3: \( B \) lying on the line of reflection (\( \overleftrightarrow{EG} \)) doesn't create a right angle in \( \triangle BCC' \) by itself, so this is incorrect.
• Option 4: By the property of reflections, the line of reflection (\( \overleftrightarrow{EG} \)) is perpendicular to the segment joining a point and its image (\( \overline{CC'} \)). So \( \overleftrightarrow{EG} \perp \overline{CC'} \), meaning \( \angle BGC \) is a right angle, so \( \triangle BGC \) is a right triangle.

Answer:

D. \( \triangle BGC \) is right because \( \overleftrightarrow{EG} \perp \overline{CC'} \).