Question

1. match.

given: \\( \angle d e f \\); point \\( i \\) in the interior of the angle;
\\( m \angle d e f = 46 ^ { \circ } ; i g = i h = 5 \mathrm { in } ; \overline { i g } \perp \overline { e g } ; \overline { i h } \perp \overline { e h } \\).
what is the measure of \\( \angle d e i \\)?
by which theorem?

Explanation:

Step1: Recall the Angle Bisector Theorem

The Angle Bisector Theorem states that if a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector. Here, \(IG = IH = 5\) in, \(IG\perp EG\), and \(IH\perp EH\). So, \(EI\) is the angle bisector of \(\angle DEF\).

Step2: Calculate the measure of \(\angle DEI\)

Since \(EI\) is the angle bisector of \(\angle DEF\) and \(m\angle DEF=46^{\circ}\), then \(m\angle DEI=\frac{1}{2}m\angle DEF\). Substituting the value of \(m\angle DEF\), we get \(m\angle DEI = \frac{1}{2}\times46^{\circ}\).

Answer:

The measure of \(\angle DEI\) is \(23^{\circ}\). By the Angle Bisector Theorem.