Question

use the figure at the right to complete the statement. all points on \\( \overrightarrow{ae} \\) are equidistant from the sides of ______. all points on \\( \overrightarrow{ae} \\) are equidistant from the sides of \\( \boxed{\blacktriangle} \\) (figure shows a triangle with point a at the top, ae as a segment from a to the base, right angles at d, g, f, b, and lengths 12 and 15 marked).

Explanation:

Step1: Recall Angle Bisector Theorem

The angle bisector theorem states that any point on the angle bisector is equidistant from the two sides of the angle.

Step2: Analyze the Figure

In the given triangle, \( \overrightarrow{AE} \) divides \( \angle BAC \) into two equal angles (as indicated by the congruent angle marks) and the perpendicular segments from points on \( AE \) to the sides \( AB \) and \( AC \) are equal (e.g., \( DF = DG \) with right angles). So, \( \overrightarrow{AE} \) is the angle bisector of \( \angle BAC \). Thus, all points on \( \overrightarrow{AE} \) are equidistant from the sides of \( \angle BAC \) (i.e., sides \( AB \) and \( AC \)).

Answer:

\( \angle BAC \) (or "the angle \( \angle BAC \)", or "sides \( AB \) and \( AC \) of \( \angle BAC \)")