QUESTION IMAGE
Question
ray ce is the angle bisector of ∠acd. which statement about the figure must be true?
- ( mangle ecd = \frac{1}{2} mangle ecb )
- ( mangle ace = \frac{1}{2} mangle acd )
- ( angle ace > angle dcb )
- ( angle ecd > angle acd )
Step1: Recall Angle Bisector Definition
An angle bisector divides an angle into two equal - measure angles. So, if ray \(CE\) is the angle bisector of \(\angle ACD\), then \(\angle ACE=\angle ECD\) and \(m\angle ACE = m\angle ECD=\frac{1}{2}m\angle ACD\).
Step2: Analyze Each Option
- Option 1: \(m\angle ECD=\frac{1}{2}m\angle ECB\). There is no information to suggest this relationship. The angle bisector is for \(\angle ACD\), not related to \(\angle ECB\) in this way.
- Option 2: \(m\angle ACE=\frac{1}{2}m\angle ACD\). Since \(CE\) bisects \(\angle ACD\), \(\angle ACE=\angle ECD\) and \(m\angle ACD=m\angle ACE + m\angle ECD = 2m\angle ACE\), so \(m\angle ACE=\frac{1}{2}m\angle ACD\). This is true.
- Option 3: \(\angle ACE>\angle DCB\). We don't know the measure of \(\angle DCB\) relative to \(\angle ACE\). There is no basis for this inequality.
- Option 4: \(\angle ECD>\angle ACD\). Since \(\angle ECD\) is a part of \(\angle ACD\) (because \(CE\) is inside \(\angle ACD\) as it is a bisector), \(m\angle ECD
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\(m\angle ACE=\frac{1}{2}m\angle ACD\) (the second option)