Question

raj correctly determined that ray lh is the bisector of ∠gli. which information could he have used to determine this? ∠glh ≅ ∠ilm m∠klm = 5m∠ilm m∠gli = 2m∠glh m∠gli = 1/2 m∠glh + 1/2 m∠hli

Explanation:

Step1: Recall the definition of an angle bisector

An angle bisector divides an angle into two equal - sized angles. If \(LH\) is the bisector of \(\angle GLI\), then \(\angle GLH=\angle HLI\) and \(m\angle GLI = m\angle GLH+m\angle HLI\). Since \(\angle GLH=\angle HLI\), we can substitute \(m\angle HLI\) with \(m\angle GLH\).

Step2: Substitute and simplify

Substituting \(m\angle HLI = m\angle GLH\) into \(m\angle GLI=m\angle GLH + m\angle HLI\), we get \(m\angle GLI=m\angle GLH+m\angle GLH\). So, \(m\angle GLI = 2m\angle GLH\).

• For \(\angle GLH\cong\angle ILM\), there is no relation to the bisector of \(\angle GLI\) as \(\angle ILM\) is not part of \(\angle GLI\).
• For \(m\angle KLM = 5m\angle ILM\), \(\angle KLM\) is a different angle (a straight - line related angle in this case) and has no direct relation to the bisector of \(\angle GLI\).
• For \(m\angle GLI=\frac{1}{2}m\angle GLH+\frac{1}{2}m\angle HLI\), if \(LH\) is the bisector \(m\angle GLI=m\angle GLH + m\angle HLI=2m\angle GLH\) (not the given formula).

Answer:

\(m\angle GLI = 2m\angle GLH\)